Slide 1

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 2

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 3

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 4

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 5

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 6

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 7

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 8

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 9

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 10

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 11

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 12

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 13

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 14

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 15

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 16

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 17

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 18

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 19

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 20

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 21

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 22

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 23

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 24

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 25

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 26

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 27

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 28

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 29

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 30

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 31

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 32

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 33

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 34

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 35

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 36

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 37

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 38

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 39

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 40

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 41

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 42

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 43

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 44

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 45

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r


Slide 46

Work and Energy Problems

Multiple Choice
A force F of strength 20N acts on an object of mass 3kg as it moves a distance
of 4m. If F is perpendicular to the 4m displacement, the work done is equal to:
a)
b)
c)
d)
e)

0J
60J
80J
600J
2400J

Since the Force is perpendicular
to the displacement, there is no
work done by this force.

Multiple Choice
Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to
6 m/s in 8s. How much work was done on the object during this time?
a)
b)
c)
d)
e)

27J
54J
72J
96J
Can not be determined from the information given
This is a job for the work energy
theorem.
W  K


1
2

m  v f  vi
2

2



  m 2  m 2 
  4 kg    6    3  
 s 
2
 s  

1

 54 J

Multiple Choice
A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in gravitational potential energy?
a)
b)
c)
d)
e)

-mgL
-mgh
-mgL/h
-mgh/L
-mghL
The length of the ramp is irrelevant,
it is only the change in height

Multiple Choice
An object of mass m is travelling at constant speed v in a circular path of radius
r. how much work is done by the centripetal force during one-half of a
revolution?

a)
b)
c)
d)
e)

πmv2 J
2πmv2 J
0J
πmv2r J
2πmv2/r J
Since centripetal force always points
along a radius towards the centre of the
circle, and the displacement is always
along the path of the circle, no work is
done.

Multiple Choice
While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above
the floor, how much work does the gravitational force do on the block?
a)
b)
c)
d)
e)

-30J
-15J
0J
15J
30J
The gravitational force points downward,
while the displacement if upward

W   m gh
m 

   2 kg   9.8 2  1.5 m 
s 

  30 J

Multiple Choice
A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that
makes an angle of 30o with the horizontal. If the block is released from rest at
the top of the incline, what is the speed at the bottom?

a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s

W  K
m gh 

1
2

gh 

1
2

The work done by gravity is
equal to the change in Kinetic
Energy.

v


m  v f  vi
2

2



2

vf

2 gh
m 

2  9.8 2  sin  30    6 m 
s 


 7.7

m
s

Multiple Choice
A block of mass 3.5kg slides down an inclined plane of length 6m that makes an
angle of 60o with the horizontal. The coefficient of kinetic friction between the
block and the incline is 0.3. If the block is released from rest at the top of the
incline, what is the speed at the bottom?
a)
b)
c)
d)
e)

4.9 m/s
5.2 m/s
6.4 m/s
7.7 m/s
9.2 m/s
m g  L sin  

Ki Ui W f  K
0  m gh  F f L 

     m g cos  

 L 
vf 

This is a conservation of
Mechanical Energy, including
the negative work done by
friction.



1
2
1
2

f

U

f

mv f  0
2

2

mv f

2 gL  sin      cos  



m 

2  9.8 2   6 m   sin  60    0.3 cos  60   
s 


 9 .2

m
s

Multiple Choice
An astronaut drops a rock from the top of a crater on the Moon. When the rock
is halfway down to the bottom of the crater, its speed is what fraction of its final
impact speed?

a)
b)
c)
d)
e)

1/4 2
1/4
1/2 2
1/2
1/ 2
Total energy is conserved. Since the rock has half
of its potential energy, its kinetic energy at the
halfway point is half of its kinetic energy at impact.

K v  v
2

1
2

Multiple Choice
A force of 200N is required to keep an object at a constant speed of 2 m/s
across a rough floor. How much power is being expended to maintain this
motion?

a)
b)
c)
d)
e)

50W
100W
200W
400W
Cannot be determined with given information
Use the power equation

P  Fv
 m
  200 N   2 
 s 
 400W

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
lifting
10.0m

300

Understanding

Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m

lifting
10.0m
300

W  F d
 m gh
m 

  2.0 kg   9.8 2  10.0 m 
s 

 196 J
 200 J

Understanding
Compare the work done on an object of mass 2.0 kg
a) In lifting an object 10.0m
b) Pushing it up a ramp inclined at 300 to the same final height

pushing
10.0m
300

The distance travelled up the ramp
d 

10.0 m
sin  30  

 20 m

W  Fapplied  d
W  m g sin    d

Applied Force
Fa p p lied  m g sin  

Work



m 

W   2.0 kg   9.8 2  sin  30    20 m 
s 

W  200 J

Example 0

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?
b) How much work is done by the normal force?
c) How much work is done by friction?
d) What is the total work done?

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

a) How much work is done by gravity?

Recall Work = force x
distance with the
force being parallel
to the distance, that
is, the component
down the ramp

W g ra vity  F  d
  m g sin     d
m 

  35 kg   9.8 2  sin  40    8 m 
s 

 1760 J

Fg

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

b) How much work is done by the normal force?
Since the normal force
is perpendicular to the
displacement, NO
WORK IS DONE.

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

c) How much work is done by friction?
Recall Work = force x distance.
W friction   F f  d
   u k m g cos  

 d

m 

   0.3   35 kg   9.8 2  cos  40    8 m 
s 

  630 J

Solution

A box slides down an inclined plane (incline angle = 400). The mass of the
block, m, is 35 kg, the coefficient of kinetic friction between the box and the
ramp, uk, is 0.3, and the length of the ramp, d, is 8m.

d) What is the total work done?
Total Work = sum of all works

W total  W gravity  W N orm al  W friction
 1760 J  0 J  630 J
 1130 J

Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

d 

1
2

v

0

 v f  t

1
m
m
  10
 0  6s
2
s
s 
 30m

Solution to Question
A truck moves with velocity v0= 10 m/s on a slick road when the driver
applies the brakes. The wheels slide and it takes the car 6 seconds to
stop with a constant deceleration.
a) How far does the truck travel before stopping?
b) Determine the kinetic friction between the truck and the road.

Fa  F f

a 
First we need the
acceleration the
car experiences.

v

m a  u Fn

t

m a  um g

0


m

 10

s

s
6s

  1 .6

m

m
s

2

Since the only
acceleration force
experienced by the
car is friction

a  ug
u 

a
g



 1 .6
 9 .8

 0 .1 7

Question
You and your bicycle have a combined mass of 80.0kg. When you reach
the base of an bridge, you are traveling along the road at 5.00 m/s. at the
top of the bridge, you have climbed a vertical distance of 5.20m and
have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the
pedals?

Solution to Question
You and your bicycle have a combined mass of 80.0kg. When you reach the base of an
bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed
a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction:
a) How much work have you done with the force you apply to the pedals?

Conservation of energy states that initial kinetic
energy equals final kinetic energy plus work done
by gravity less work done by you
W   ET
 K f U
Wp  K


1
2

f

f

 K

 Ki U

mv f 
2

1
2

f

i

Ui

Ui

m v i  m gh  0
2

2
2

m
m
m 

 

  80.0 kg    1.50    5.00     80 kg   9.8 2   5.2 m 
2
s 
s  
s 


 

1

 3166.8 J
 3.17  10 J
3

Question
The figure below shows a ballistic pendulum, the system for measuring the
speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of
M=3.0kg , suspended like a pendulum, and makes a completely inelastic
collision with it. After the impact of the bullet, the block swings up to a maximum
height 2.00 cm. Determine the initial velocity of the bullet?
Stage 1 (conservation of Momentum)
pi  p f
m v1i  M v 2 i  m V  M V

 0.001kg  v1  0   3.001.kg  V
v1  3001V

Stage 2 (conservation of Energy)
Ki  U
1
2

 m  M V 2
V

2

f

 m  M

 gh

m 

 2  9.8 2   0.02 m 
s 


V  0.626

m
s

Therefore:

m

v1  3001  0.626 
s 

 1879

m
s

Question
We have previously used the following expression for the maximum
height h of a projectile launched with initial speed v0 at initial angle θ:
2

h

v 0 sin

2

 

2g
Derive this expression using energy considerations

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

This initially appears to be an easy
problem: the potential energy at point
2 is U2=mgh, so it may seem that all
we need to do is solve the energyconservation equation K1+U1=K2+U2
for U2. However, while we know the
initial kinetic energy and potential
energies (K1=½mv12=½mv02 and
U1=0), we don’t know the speed or
kinetic energy at point 2. We will need
to break down our known values into
horizontal and vertical components
and apply our knowledge of kinematics
to help.

2

h

v 0 sin

2

2g

 

Solution to Question
We have previously used the following expression for the maximum height h of a projectile
launched with initial speed v0 at initial angle θ:
Derive this expression using energy considerations

We can express the kinetic energy at
each point in the terms of the
2
2
2
components using: v  v x  v y
K1 

1

K2 

1

2
2

m  v1 x  v1 y 
2

2

m  v2 x  v2 y 
2

2

Conservation of energy then gives:
 2 gh
y K U
K 1  Uv11 
2
2
2

1
2

m v

2
1x

v sin
   1 2 gh 2
2
2
 0v1 y   0  m  v 2 x  v 2 y   m gh
gh
2
2
2 v sin   
0
2
2 h 2
2
v1 x  v1 y  v22gxy h v222ggy h 2 gh
2

2

But, we recall that
Furthermore,
the x-components
But,
v1y is just the ysince
projectile
of velocity
does
not
component
of initial
has
zero
vertical
change, vv1y
=v02xsin(θ)
velocity:
1x=v
velocity at highest
points, v2y=0

2

h

v 0 sin

2

2g

 

Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg)
moves through a quarter-circle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the
curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the
bottom is only 6.00 m/s. what work was done by the friction force
acting on him?

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

From Conservation of Energy

K1  U 1  K 2  U 2
0  m gR 

1
2

v2 


m v2  0
2

2 gR
m 

2  9.8 2   3.00 m 
s 


 7.67

m
s

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The free body diagram of the normal
is through-out his journey is:

F

y

 m ac
2

F N  FG  m

v2
R

 2 gR 
FN  m g  m 

 R 
 mg  2mg
 3m g

v2 

2 gR

Solution to Question
Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m.
a) Determine his speed at the bottom of the ramp
b) Determine the normal force that acts on him at the bottom of the curve.
c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00
m/s. what work was done by the friction force acting on him?

The normal force does no work, but
the friction force does do work.
Therefore the non-gravitational work
done on Vinney is just the work done
by friction.

The free body diagram of the normal
is through-out his journey is:

K1  U 1  WF  K 2  U 2
WF  K 2  U 2  K1  U 1


1
2

m v 2  0  0  m gh
2

2

m
m 


  25.0 kg   6.00    25.0 kg   9.80 2   3.00 m 
2
s 
s 


1

 285 J

Example
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.

1.0 kg

Solution
An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the
coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine
the distance the object travels before coming to a halt.
FN
Forces
We can see that the objects weight is
balanced by the normal force exerted by
the ice. Therefore the only work done is
due to the friction acting on the object.
Let’s determine the friction force.
F f  u k FN
 uk mg

m 

  0.10  1.0 kg   9.8 2 
s 

 0 .9 8 N

Ff

W  KE
Ff d 

  0.98 N  d



d 

Now apply the work Energy
Theorem and solve for d

1

mv 

2

1

2
f

1

mg
mv

2



1.0 kg   0

2

 12.5 J

 0.98 N

 1 3m

2
i

2

m
1
m


1.0
kg
5.0





s 
2
s 


2

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.

A

a)
b)
c)
d)
e)

Find the speed of the box when its height above the ground is 1/2H
Find the speed of the box when it reaches B.
Determine the value of uk, so that it comes to a rest at C
Determine the value of uk if C was at a height of h above the ground.
If the slide was not frictionless, determine the work done by friction as
the box moved from A to B if the speed at B was ½ of the speed
calculated in b)

H

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
a) Find the speed of the box when its height above the ground is 1/2H

E total  U G  K  m gH  0  m gH
A

U
G

1

 K 1  E total

2

2

1
 1
2
m g  H   m v  m gH
2
 2

1

H

mv 
2

2

1

m gH

2
v

gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
b) Find the speed of the box when it reaches B.

E total  U G  K  m gH  0  m gH
A

U G B  K B  E total
0

1
2

H

m v  m gH
2

v  2 gH
2

v

2 gH

uk
B

x

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
c) Determine the value of uk, so that it comes to a rest at C

A

H

W  K
1
1
2
2
W  m v f  m vi
2
2
1
2
 m vi
2
1
2
F d  m vi
2
1
2
m gu k x  m v i
2

v

2

uk 


x

vi

2 gx
H
x

uk
B

2 gH

C

Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
d) Determine the value of uk if C was at a height of h above the ground.

KB U B W f  KC UC
m gH  0  F f L  0  m gh
A

m g H  u k m g co s    L  m g h

H

uk 

H  u k co s    L  h



H h
L cos  
H h
x

L
B

uk

x

C



Example
A box of mass m is released from rest at point A, the top of a long frictionless
slide. Point A is at height H above the level points B and C. Although the slide is
frictionless, the horizontal surface from point B to C is not. The coefficient of
kinetic friction between the box and this surface is uk, and the horizontal
distance between points B and C is x.
e) If the slide was not frictionless, determine the work done by friction as the box
moved from A to B if the speed at B was ½ of the speed calculated in b)

A

H
uk

U A  KA Wf UB  KB
1
2
U A  K A  W f  m vB
2
2
1 1

m gH  0  W f  m 
2 gH 
2 2

m gH
m gH  W f 
4
m gH
3
Wf 
 m gH   m gH
4
4
B

x

Example
An acrobat swings from the horizontal. When the acrobat was swung an angle
of 300, what is his velocity at that point, if the length of the rope is L?
Because gravity is a
conservative force (the
work done by the rope is
tangent to the motion of
travel), Gravitational
Potential Energy is
converted to Kinetic
Energy.
Ui  Ki U
m gL 

1
2

1

m gL 

2

1

v

f

m v  m gL sin  30  
2

30

mv

2
gL  v

It’s too difficult to
use Centripetal
Acceleration

2

gL

2

L

h  L sin  3 0  

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C
b) Determine the speed of the car when its position relative to B is specified by the
angle θ shown in the diagram.
c) What is the minimum cut-off speed vc that the car must have at D to make it
around the loop?
d) What is the minimum height H necessary to ensure that the car makes it around
the loop?
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
A
D

H
E

θ
B

C
F

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
a) Find the centripetal acceleration of the car when it is at Point C

A

K A  U A  KC  U c

D

H

E

θ
B

0  m gH 

C

1
2

F

m v C  m gr
2

vC  2 g  H  r 
2

2

vC

The centripetal
acceleration is v2/r



r

aC 

2g H  r 
r

2g H  r
r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
b) Determine the speed of the car when its position relative to B is specified by the
A angle θ shown in the diagram.
KA U A  K U
D
1
2
H
0  m gH  m v  m g  r  r cos 180   
E
2
θ C
F
B
0  m gH 

The car’s height above the
bottom of the track is given by
h=r+rcos(1800-θ).

1
2



m v  m g  r  r cos  
2

m g H  r 1  cos  
v

 

1

mv

2

2 g  H  r  1  co s  


  

2




Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
c) What is the minimum cut-off speed vc that the car must have at D to make it
A around the loop?

D

H

E

θ
B

FC  FG  FN

C
F
m

When the car reaches D, the
forces acting on the car are its
weight, mg, and the
downward Normal Force.
These force just match the
Centripetal Force with the
Normal equalling zero at the
cut-off velocity.

v

2

 mg  0

r
v cu t  o ff 


rm g
m
gr

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
d) What is the minimum height H necessary to ensure that the car makes it around
A the loop?
KA U A  KD UD

D

H

E

C
B

0  m gH 

1
2

F
m gH 

1
2

Apply the cut-off speed from c)
to the conservation of
Mechanical Energy.



5

mv  mg  2r 
2

m  gr   m g  2 r 

m gr

2

H 

5
2

r

Example
The diagram below shows a roller-coaster ride which contains a circular loop of
radius r. A car (mass m) begins from rest from Point A and moves down a
frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling
once around the circle (B to C to D to E back to B) it then travels along a flat portion
from B to F (which is not frictionless).
e) If H=6r and the coefficient of friction between the car and the flat portion of the
track from B to F is 0.5, how far along the flat portion of the track will the car
travel before coming to rest at F?
KA U A  KB UB
A
1
2
0

m
g
(6
r
)

m
v
0
B
D
2
H
E
C
F
B
F f x  6 m gr

Calculate the Kinetic Energy at
B, then determine how much
work friction must do to
remove this energy.

 k mgx  6 mgr
x

6r

k



6r
0.5

 12 r