Kinetic Energy and Work

Transcript Kinetic Energy and Work

Chapter 7 Kinetic Energy and Work

Introduction to Energy

 The concept of energy is one of the most important topics in science  Every physical process that occurs in the Universe involves energy and energy transfers or transformations

Work and Energy

Energy: scalar quantity associated with a state (or condition) of one or more objects. Work and energy are scalars, measured in N ·m or Joules, J, 1J=kg ∙m 2 /s 2 Energy can exist in many forms - mechanical, electrical, nuclear, thermal, chemical….

Energy Approach to Problems

 The energy approach to describing motion is particularly useful when the force is not constant  An approach will involve

Conservation of Energy

• This could be extended to biological organisms, technological systems and engineering situations

Work and Energy

Energy is a conserved quantity - the total amount of energy in the universe is constant.

Energy can be converted from one type to another but never destroyed.

Work and energy concepts can simplify solutions of mechanical problems - they can be used in an alternative analysis

Systems

 A

system

is a small portion of the Universe • We will ignore the details of the rest of the Universe  A critical skill is to identify the system

Valid System

 A valid system may • • • • be a single object or particle be a collection of objects or particles be a region of space vary in size and shape

Environment

 There is a

system boundary

around the system • • The boundary is an imaginary surface It does not necessarily correspond to a physical boundary  The boundary divides the system from the

environment

• The environment is the rest of the Universe

Work

 The work,

, done on a system by an agent exerting a constant force on the system is the product of the magnitude, of the force, the magnitude

Δr

of the

, displacement of the point of application of the force, and

cos

where

is the angle between the force and the displacement vectors

Work



W = F Δr cos θ

• • • The displacement is that of the point of application of the force A force does no work on the object if the force does not move through a displacement The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application

Work Example

 The normal force,

, and the gravitational force,

, do no work on the object

cos θ = cos 90° = 0

 The force

does do work on the object

More About Work

 The system and the environment must be determined when dealing with work  The environment does work on the system (Work

the environment

the system)  The sign of the work depends on the direction of

relative to

Δr

 Work is positive when projection of

onto

Δr

is in the same direction as the displacement  Work is negative when the projection is in the opposite direction

Work Is An Energy Transfer

• This is important for a system approach to solving a problem • If the work is done on a system and it is positive, energy is transferred to the system • If the work done on the system is negative, energy is transferred from the system

Work done by a constant force

Work done on an object by a constant force is defined to be the product of the magnitude of the

displacement

and the component of the

force

parallel to the displacement

W = F II · d

Where

F II

is the component of the force parallel to the displacement

d F

Work

In other words -

W = F d cos

q Where

is the angle between

and

d F

is > 90 o , work is negative. A decelerating car has negative work done on it by its engine.

The unit of work is called

Joule (J) , 1 J = 1 N ·m

1J=kg ∙m

/s

Scalar Product of Two Vectors

 The scalar product of two vectors is written as

A . B

 It is also called the dot product

A . B = A B cos

θ θ

is the angle

between

and

Scalar Product

 The scalar product is commutative

A . B = B . A

 The scalar product obeys the distributive law of multiplication

A . (B + C)

A . B + A . C

Dot Products of Unit Vectors

 iˆ  iˆ  jˆ  jˆ  kˆ  kˆ  1 iˆ  jˆ  iˆ  kˆ  jˆ  kˆ  0  Using component form with

and

: A  A x iˆ  A y jˆ  A z kˆ B  B x iˆ  B y jˆ  B z kˆ A  B  A x B x  A y B y  A z B z

Work - on and by

A person pushes block

30 m

along the ground by exerting force of

25 N

on the trolley. How much work does the person do on the trolley?

W = F d = 25N x 30m = 750 Nm

Trolley does -

750 Nm

work on the person

F tp F p

  4

ˆ  3

ˆ  N the object moves in the

direction from the origin to

x = 5.00 m

. Find the work

W =





done on the object by the force.

  4

ˆ  3

ˆ  N the object moves in the

direction from the origin to

x = 5.00 m

. Find the work

W =





done on the object by the force.

 

f i

  5 m 0   4

€  3

 N 

5 m 0   4 N m 

xdx

0  4 N m 

2 2 0 5 m  50.0 J

Mechanical Energy

Mechanical energy (energy associated with masses) can be thought of as having two components:

kinetic

and

potential

– Kinetic energy is energy of motion – Potential energy is energy of position

Kinetic Energy

In many situations, energy can be considered as “the ability to do work” Energy can exist in different forms Kinetic energy is the energy associated with the

motion

of an object A moving object can do work on another object – E.g hammer does work on nail.

Kinetic Energy

Consider an object with mass

moving in a straight line with initial velocity uniformly to a speed

v f v i

. To accelerate it a constant net force

is exerted on it parallel to motion over a distance



2  2

(



) 

2  2

Work done on object



 2

W = F d = m a d

(NII) So



 2

i d

Kinetic Energy

If we rearrange this we obtain

 1 2

 1 2

2 We define the quantity

½mv 2

to be the translational kinetic energy (KE) of the object

W =

This is the ‘Work-Energy Theorem’:

“The net work done on an object is equal to its change in kinetic energy ”

The Work-Energy Theorem

W =

• Note – The net work done on an object is the work done by the net force.

– Units of energy and work must be the same (J)

Energy:

scalar quantity associated with a state (or condition) of one or more objects. I.

Kinetic energy.

II. Work.

III. Work - Kinetic energy theorem.

IV. Work done by a constant force - Gravitational force V.

Work done by a variable force.

- Spring force.

1D-Analysis 3D-Analysis VI.

Power

I. Kinetic energy Energy associated with the state of motion of an object.

 1 2

2 Units:

1 Joule = 1J = 1 kgm 2 /s 2

II. Work Energy transferred “to” or “from” an object by means of a force acting on the object.

To  From +W  -W

- Constant force :

F x



ma x v



2 0  2

a x d



a x



 2

d v

2 0  1 2

(



2 0 ) 

K f



K i

 

F x



ma x

 1 2

(



2 0 ) 1

d F x d





F x d

Work done by the force = Energy transfer due to the force.

-To calculate the work done on an object by a force during a displacement, we use only the force component along the object’s displacement. The force component perpendicular to the displacement does zero work.



F x d



cos   

 

Assumptions:

1) F = constant force 2) Object is particle-like (rigid object, all parts of the object must move together).

A force does

when it has a vector component in the same direction as the displacement, and

–W

when it has a vector component in the opposite direction.

has no such vector component.

W=0

when it   90   

180     90   

  90   0

Net work done by several forces = Sum of works done by individual forces.

Calculation:

1) W net = W 1 +W 2 +W 3 + … 2) F net  W net =F net d

II. Work-Kinetic Energy Theorem



K f



K i



Change in the kinetic energy of the particle = Net work done on the particle

III. Work done by a constant force

- Gravitational force:



mgd

cos  Rising object:

F g

transfers

W= mgd cos180 º = -mgd mgd

energy

from

the  object’s kinetic energy.

Falling down object:

F g

transfers

mgd W= mgd cos 0 º = +mgd

energy

the  object’s kinetic energy .

External applied force + Gravitational force:

D



K f



K i



W a



W g

Object stationary before and after the lift:

W a +W g =0

The applied force transfers the same amount of energy to the object as the gravitational force transfers from the object.

IV. Work done by a variable force

- Spring force: 

 



Hooke’s law k

= spring constant  measures spring’s stiffness.

Units:

N/m

Hooke’s Law

 When

is positive (spring is stretched), negative

is  When

(at the equilibrium position),

is  When

is negative (spring is compressed),

is positive

Hooke’s Law

 The force exerted by the spring is always directed opposite to the displacement from equilibrium  

is called the

restoring force

If the block is released it will oscillate back and forth between

–x

and

Hooke’s law:



F x

 

x i Δx x Work done by a spring force: Assumptions: • Spring is massless • Ideal spring   obeys • Contact between the block and floor is frictionless.

• Block is particle-like.

m spring << m block

Hooke’s law exactly.

- Calculation: 1) The block displacement must be divided 2) into many segments of infinitesimal width,

F(x) ≈ cte

within each short

Δx

segment.

Δx

F x x i F j Δx x f x

W s

 

F j

 D



x i x f F dx

 

x i x f

( 

)

 



x i x f W s

 1 2

k x i

2  1 2

k x

f W s

  1 2

k x

f if



x i

 0 0 

x dx

    1 2

    

x x i f

    1 2

   (



x i

2 ) W s >0 relaxed position (x=0) than it was initially.

W s <0  Block ends up further away from x=0 .

 Block ends up closer to the W s =0  Block ends up at x=0 .

Work done by an applied force + spring force: D



K f



K i



W a



W s

Block stationary before and after the displacement:

W a = -W s

 The work done by the

applied force

displacing the block is the negative of the work done by the

spring force

If it takes law spring

4.00 J

of work to stretch a Hooke's-

10.0 cm

from its unstressed length, determine the extra work required to stretch it an additional

10.0 cm

If it takes law spring

4.00 J

of work to stretch a Hooke's-

10.0 cm

from its unstressed length, determine the extra work required to stretch it an additional

10.0 cm

4.00 J



1 2



0.100 m

 2

800 N m 1 2

 2 

4.00 J



12.0 J

2.00-kg

block is attached to a spring of force constant

500 N/m.

The block is pulled

5.00 cm

to the right of equilibrium and released from rest. Find the speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface is

0.350

2.00-kg

block is attached to a spring of force constant

500 N/m.

The block is pulled

5.00 cm

0.350

W

v f s s

 



1 2

 1 2

 1   2 1 2 2 

m v









1 2

m v

2.00



  1 2

m v

m s   2



0  2 0.791 m s

2.00-kg

block is attached to a spring of force constant

500 N/m.

The block is pulled

5.00 cm

0.350

1 2

m v

i s



1 2

m v

0         1 2

m v

0.282 J 

v f

 1  2 2.00 kg  2.00



m s  0.531 m s

Work done by a general variable force:

• Assume that during a very small displacement,

Δx

is constant • For that displacement,

W ~ F Δx

• For all of the intervals,





f x i x

Work done by a general variable force:

• lim



f x i x

D  

x i x f F dx x

• Therefore,

 

x i x f F dx x

• The work done is equal to the area under the curve

3D-Analysis



F d r

 

F x



dx i

 ˆ

F y

 ˆ

j dy



F z k

ˆ 

;

F x



(

F y



(

F z



(

)

 



d r

 

F x dx



F y dy



F z dz





r r i



f dW



x x i



f F x dx



y y i



f F y dy



i z z



f F z dz W

Work-Kinetic Energy Theorem - Variable force

x x f f



x i



(

)





x i ma dx ma dx



m dv dt dx



m dv dx v dx



mvdv dv dt



dv dx dx dt



dv dx v W



v i v f



mv dv



m v f



v i v dv

 1 2

 1 2

mv i

2 

K f



K i

 D

V. Power

Time rate at which the applied force does work.

- Average power: amount of work done in an amount of time

Δt

by a force.

P avg



-Instantaneous power: -instantaneous time rate of doing work.



dW dt

Units: 1 watt= 1 W = 1J/s

1 kilowatt-hour = 1000W ·h = 1000J/s x 3600s = 3.6 x 10 6 = 3.6 MJ J F



dW dt



cos 

dx dt



cos    

dx dt

   

cos   



 φ

In the figure below a downward angle

θ 2N

force is applied to a

4kg

block at a as the block moves right-ward through across a frictionless floor. Find an expression for the speed the end of that distance if the block’s initial velocity is: (a)

0 v f

at and (b)

1m/s

to the right.

(a)

N F x F x F y N F y mg mg

In the figure below a downward angle

θ 2N

force is applied to a

4kg

block at a as the block moves right-ward through

across a frictionless floor. Find an expression for the speed the end of that distance if the block’s initial velocity is: (a)

0 vf

at and (b)

1m/s

to the right.

(a)

F x F y mg F x F y N mg

W W

 

 

 (

cos q )

 D

 0 .

(



0 2 ) (

)

0  0  D

 0 .

( 2

)( 1

) cos q  0 .

5 ( 4

)



v f

 cos q

(

)

0  1

 D

 0 .

 2

( 2

) cos q  0 .

5 ( 4

)

 2



v f

 1  cos q

1m/s

to the right, but now the

force is directed downward to the left. Find an expression for the speed of the block at the end of the 1m distance.

N F x F y mg F x F y N mg

1m/s

to the right, but now the

force is directed downward to the left. Find an expression for the speed of the block at the end of the 1m distance.

N F x F x F y N mg

W W

  

 

  (

0 .

cos q (

v f

2  )

d v

0 2 )

F y mg

(

)

0  1

 D

 0 .

 2

( 2

)( 1

) cos q  0 .

5 ( 4

)

 2



v f

 1  cos q

(

)

0  1

 D

 0 .

 2

 ( 2

) cos q  0 .

5 ( 4

)

 2



v f

 1  cos q

A small particle of mass

is pulled to the top of a frictionless half-cylinder (of radius

) by a cord that passes over the top of the cylinder, as illustrated in Figure. (a) If the particle moves at a constant speed, show that

F = mgcos

. (

Note

: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times.) (b) By directly integrating

W =





, find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder.

A small particle of mass

is pulled to the top of a frictionless half-cylinder (of radius

) by a cord that passes over the top of the cylinder, as illustrated in Figure. (a) If the particle moves at a constant speed, show that

F = mgcos

. (

Note

: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times.) (b) By directly integrating

W =





, find the work done in moving the particle at constant speed from the bottom to the top of the half cylinder. (a) 

F



m a

cos q  0

F



m g

cos q

A small particle of mass

is pulled to the top of a frictionless half-cylinder (of radius

) by a cord that passes over the top of the cylinder, as illustrated in Figure. (a) If the particle moves at a constant speed, show that

F = mgcos

. (

Note

: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times.) (b) By directly integrating

W =





, find the work done in moving the particle at constant speed from the bottom to the top of the half cylinder. (b)

W W

 

 

i f

  0



m g

cos q q

m gR

   

m gR m gR

sin q  0 2 ( We use radian measure to express the next bit of displacement dr as in terms of the next bit of angle moved through d θ : dr=Rd θ )

Two springs with negligible masses, one with spring constant

and the other with spring constant When the glider is in equilibrium, spring

, are attached to the endstops of a level air track as in Figure. A glider attached to both springs is located between them. is stretched by extension

spring force

F 2

app to the right of its unstretched length and is stretched by

to the left. Now a horizontal is applied to the glider to move it a distance

x a

the right from its equilibrium position. Show that in this process (a) the work done on spring

1 (x

2 +2x

a x

i1 )

to , (b) the work done on spring related to

x i

1 the force

app

is (

2 is

+ k 2 )x

2 (x

2 – 2x

a x

i2 )

, (c)

is by

i2 = k 1

i1 /k 2

, and (d) the total work done by

(a) (b)

Two springs with negligible masses, one with spring constant

and the other with spring constant When the glider is in equilibrium, spring

, are attached to the endstops of a level air track as in Figure. A glider attached to both springs is located between them. is stretched by extension

spring force

F 2

app to the right of its unstretched length and is stretched by

to the left. Now a horizontal is applied to the glider to move it a distance

x a

the right from its equilibrium position. Show that in this process (a) the work done on spring

1 (x

2 +2x

a x

i1 )

to , (b) the work done on spring related to

x i

1 the force

app

is (

2 is

+ k 2 )x

2 (x

2 – 2x

a x

i2 )

, (c)

is by

i2 = k 1

i1 /k 2

, and (d) the total work done by

W W

1 2  



 

x i

2   

x i

x a x i

1  

x i

x a

 1 2

1   

x i

1 

x a

 2 

1    1 2  1 2

2    

x i

2 

x a

 2 

2    1 2   2

 2

x x

1  2

 2

x x

2 

Two springs with negligible masses, one with spring constant

and the other with spring constant When the glider is in equilibrium, spring

, are attached to the endstops of a level air track as in Figure. A glider attached to both springs is located between them. is stretched by extension

spring force

F 2

app to the right of its unstretched length and is stretched by

to the left. Now a horizontal is applied to the glider to move it a distance

x a

the right from its equilibrium position. Show that in this process (a) the work done on spring

1 (x

2 +2x

a x

i1 )

to , (b) the work done on spring related to

x i

1 the force

app

is (

2 is

+ k 2 )x

2 (x

2 – 2x

a x

i2 )

, (c)

is by

i2 = k 1

i1 /k 2

, and (d) the total work done by

(c)

Before the horizontal force is applied, the springs exert equal forces:

k x



k x i

x i

2 

k x k

(d)

Two springs with negligible masses, one with spring constant

and the other with spring constant When the glider is in equilibrium, spring

, are attached to the endstops of a level air track as in Figure. A glider attached to both springs is located between them. is stretched by extension

spring force

F 2

app to the right of its unstretched length and is stretched by

to the left. Now a horizontal is applied to the glider to move it a distance

x a

the right from its equilibrium position. Show that in this process (a) the work done on spring

1 (x

2 +2x

a x

i1 )

to , (b) the work done on spring related to

x i

1 the force

app

is (

2 is

+ k 2 )x

2 (x

2 – 2x

a x

i2 )

, (c)

is by

i2 = k 1

i1 /k 2

, and (d) the total work done by

1 

 1 2 2 2

  1 2 1 2  1  2

1   2

 2



a i

1  1 2

a i

1  2



a k x k

a i

N6 . A

2kg

lunchbox is sent sliding over a frictionless surface, in the positive direction of an

axis along the surface.

Beginning at

t=0

, a steady wind pushes on the lunchbox in the negative direction of lunchbox at (a)

t=1s x

, (b) Estimate the kinetic energy of the

t=5s

. (c) How much work does the force from the wind do on the lunch box from

t=1s

t=5s

N6.

2kg

lunchbox is sent sliding over a frictionless surface, in the positive direction of an

axis along the surface. Beginning at

t=0

, a steady wind pushes on the lunchbox in the negative direction of lunchbox at (a)

t=1s

, (b)

Estimate the kinetic energy of the

t=5s

. (c) How much work does the force from the wind do on the lunch box from

Motion



t=1s

concave

t=5s

downward parabola x



 1 10



dx dt

 1  2 10

t a



dv dt

  2 10   0 .



cte



 ( 2

)(  0 .

2 )   0 .

N W





 (  0 .

)(

 0 .

2 ) (

)

 1



v f K f

 0 .

5 ( 2

)( 0 .

) 2  0 .

N6.

A 2kg lunchbox is sent sliding over a frictionless surface, in the positive direction of an x axis along the surface. Beginning at t=0, a steady wind pushes on the lunchbox in the negative direction of x, Fig. below. Estimate the kinetic energy of the lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force from the wind do on the lunch box from t=1s to t=5s?

Motion



concave downward parabola F



cte



 ( 2

)(  0 .

2 )   0 .

N W





 (  0 .

)(

 0 .

2 )



 1 10



dx dt

 1  2 10

t a



dv dt

  2 10   0 .

2 (

b K

)

f t

 5



v f

 0

(

)

 D



K f

( 5

) 

K f

( 1

)  0  0 .

64   0 .

 0

N12.

20N

In the figure below a horizontal force is applied to a

3kg F a

of magnitude book, as the book slides a distance of

d=0.5m

up a frictionless ramp. (a) During the displacement, what is the net force done on the book by

F a

, the gravitational force on the book and the normal force on the book? (b) If the book has zero kinetic energy at the start of the displacement, what is the speed at the end of the displacement?

x y

N F gx mg F gy

N12.

In the figure below a horizontal force is applied to a

3kg F a

of magnitude book, as the book slides a distance of

20N d=0.5m

F gx N F gy

up a frictionless ramp. (a) During the displacement, what is the net force done on the book by

F a

, the gravitational force on the book and the normal force on the mg book? (b) If the book has zero kinetic energy at the start of the displacement, 



 

 0 what is the speed at the end of the displacement?

Only F gx

F ax do work

(

)



W Fa x F net W net



F ax



F gx

 ( 17 .



W Fg x or

 20 cos 30   14 .

) 

sin 30  0 .

 1 .

W net

 

F net





N12.

In the figure below a horizontal force is applied to a

3kg F a

of magnitude book, as the book slides a distance of

20N d=0.5m

F gx N F gy

up a frictionless ramp. (a) During the displacement, what is the net force done on the book by

F a

, the gravitational force on the book and the normal force on the mg book? (b) If the book has zero kinetic energy at the start of the displacement, 



 

 0 what is the speed at the end of the displacement?

Only F gx

F ax do work

(

b W

) 

0  0 

 D



K f

1 .

 0 .



v f

 0 .

N15.

(a) Estimate the work done represented by the graph below in displacing the particle from by F=a/x 2 , with a=9Nm 2 x=1 to x=3m . (b) The curve is given . Calculate the work using integration

N15.

(a) Estimate the work done represented by the graph below in displacing the particle from by F=a/x 2 , with a=9Nm 2 x=1 to x=3m . (b) The curve is given . Calculate the work using integration (

)



Area under curve

 ( 11 .

squares

)( 0 .

)( 1

)  5 .

(

)

 1 3  9

   9  1

  1 3   9 ( 1 3  1 )  6

N19.

An elevator has a mass of 4500kg and can carry a maximum load of 1800kg . If the cab is moving upward at full load at constant speed 3.8m/s , what power is required of the force moving the cab to maintain that speed?

F a mg

N19.

F a

m total



F net

  

F a

4500





  1800 

m a



0  6300



F a



 ( 6300

)( 9 .

2 )  61 .

 



  ( 61 .

)( 3 .

)  234 .

N17.

A single force acts on a body that moves along an x-axis.

The figure below shows the velocity component versus time for

the body. For each of the intervals

A B C D t

AB, BC, CD, and DE, give the sign (plus or minus) of the work done by the force, or state that the work is zero.

N17.

A single force acts on a body that moves along an x-axis.

The figure below shows the velocity component versus time for

the body. For each of the intervals

A B C D t

AB, BC, CD, and DE, give the sign (plus or minus) of the work done by the force, or state that the work is zero.

 D

K AB



v B

 

K f v A

 

 1  2 0





0 2 



v C



v B



 0



v D



v C



 0



v E

 0 ,

v D

 0 

 0

15E.

In the

figure below, a cord runs around two massless, frictionless pulleys; a canister with mass pulley; and you exert a force

F m=20kg

hangs from one on the free end of the cord.

(a) What must be the magnitude of

if you are to lift the canister at a constant

speed? (b) To lift the canister by

2cm

T T T P1 mg

how far must you pull the free end of the cord? During that lift, what is the work done on the canister by (c) your force (via the cord) and (d) the gravitational force on the canister?

15E.

In the

figure below, a cord runs around two massless, frictionless pulleys; a canister with mass pulley; and you exert a force

F m=20kg

hangs from one on the free end of the cord.

(a) What must be the magnitude of

if you are to lift the canister at a constant

speed? (b) To lift the canister by

2cm

T T T P1 mg

how far must you pull the free end of the cord? During that lift, what is the work done on the canister by (c) your force (via the cord) and (d) the gravitational force on the canister?

(

)

Pulley

1 :



cte



F net

 0  2



 0 

 98

N Hand



cord



 0 



2  98

15E.

In the

figure below, a cord runs around two massless, frictionless pulleys; a canister with mass pulley; and you exert a force F m=20kg hangs from one on the free end of the cord.

(a) What must be the magnitude of F if you are to lift the canister at a constant

speed? (b) To lift the canister by 2cm ,

T T T P1 mg

how far must you pull the free end of the cord? During that lift, what is the work done on the canister by (c) your force (via the cord) and (d) the gravitational force on the canister?

(b) To rise “m” 0.02m

, two segments of the cord must be shorten by that amount. Thus, the amount of the string pulled down at the left end is: 0.04m

(

)

W F





 ( 98

)( 0 .

)  3 .

15E.

In the

figure below, a cord runs around two massless, frictionless pulleys; a canister with mass m=20kg hangs from one pulley; and you exert a force F on the free end of the cord.

(a) What must be the magnitude of F if you are to lift the canister at a constant speed? (b) To lift the canister by 2cm,

T T T P1 mg

how far must you pull the free end of the cord? During that lift, what is the work done on the canister by (c) your force (via the cord) and (d) the gravitational force on the canister?

(

)

W Fg

 

mgd

 (  0 .

)( 20

)( 9 .

2 )   3 .

W F +W Fg =0 There is no change in kinetic energy.

Challenging problems – Chapter 7

Two trolleys of masses m 1 =400 kg and m 2 =200 kg are connected by a rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes to push them with a force of 1200N.

From the point of view of kinetics, does the relative position of the

Situation 1 N 1 F 1r F 2r N 2 F

trolleys matter? If the rod can only stand an applied force of 500N,

m 1 g m 2 g

which trolley should be up front?

Challenging problems – Chapter 7

Two trolleys of masses m 1 =400 kg and m 2 =200 kg are connected by a rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes to push them with a force of 1200N.

From the point of view of kinetics, does the relative position of the

Situation 1 N 1 F 1r F 2r N 2 F

trolleys matter? If the rod can only stand an applied force of 500N,

m 1 g m 2 g

which trolley should be up front?

F’ 2r Situation 1

m m

1 2 : :

F F

2 

r F

 1

F’ 1r

 2

a m

Action and reaction forces: F 1r F 2r force that the rod does on m 1 force that the rod does on m 2 F’ 1r F’ 2r force that m 1 does on rod.

force that m 2 does on rod.



a F

 1

(

1 

2 

r F



2  ) 400

2 

1200

 200



 400

Rod



negiglible mass, F net on rod=0 F’ 1r = F’ 2r



rigid rod does not deform F 1r = F’ 1r



Action / Reaction

Forces that the rod does on the blocks as it tries to counteract the deformation that F could induce.

Challenging problems – Chapter 7

Two trolleys of masses m 1 =400 kg and m 2 =200 kg are connected by a rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes to push them with a force of 1200N. From the point of view of kinetics, does the relative position of the trolleys matter? If the rod can only stand an applied force of 500N, which trolley should be up front?

F’ 2r Situation 2 F’ 1r Action and reaction forces:

2 :

1 :

F F







  1200

r N



 400

 200

a a F

 1

(

1 





2 )  800

F’ Rod



1r F 1r F 1r F 2r F’ 1r F’ 2r force that the rod does on m 1 force that the rod does on m 2 force that m 1 does on rod.

force that m 2 does on rod.

negiglible mass, F net = F’ 2r



= F’ 1r



Action / Reaction Situation 2

From kinetic point Situation1=Situation2  same of view, acceleration”.

From dynamics  F 1r (1 )≠F 1r (2) Only situation (1) is possible F 1r =F’ 1r = 400N

F m 2 g N 2 F 2r F 1r on rod=0 rigid rod does not deform m 1 g N 1

A car with a weight of develops a velocity of 2500N working with a power of 112 km/hour 130kW when traveling along a horizontal straight highway. Assuming that the frictional forces (from the ground and air) acting on the car are constant but not negligible: What is the value of the frictional forces?

(a) What is the car’s maximum velocity on a 5 0 incline hill? (b) What is the power if the car is traveling on a 10 0 inclined hill at 36km/h ?

A car with a weight of 2500N working with a power of 130kW develops a velocity of 112 km/hour when traveling along a horizontal straight highway. Assuming that the frictional forces (from the ground and air) acting on the car are constant but not negligible: What is the value of the frictional forces?

(a) What is the car’s maximum velocity on a 5 0 incline hill? (b) What is the power if the car is traveling on a 10 0 inclined hill at 36km/h?

N f F Situation 1 mg

 130

kW v

 112

h P

 

 



 31 .



P v

 130 31 .

kW m

 4178 .

N F





 0 (



cte

) 



 4178 .

Situation 2

 

F mg

sin q   4396 .

N f

 0 

 ( 2500

) sin 5   4178 .

 0 (

)



P F

 130000

4396 .

 29 .

f N F gx F gy F mg

(

)

 36

 10

, q  10 

 ( 2500

) sin 10   4178 .

N P





 46128 .

 0 

 4612 .

Kinetic Energy and Work

Transcript Kinetic Energy and Work

Chapter 7 Kinetic Energy and Work

Introduction to Energy

Work and Energy

Energy Approach to Problems

Work and Energy

Systems

Valid System

Environment

Work

Work

Work Example

More About Work

Work Is An Energy Transfer

Work done by a constant force

Work

1J=kg ∙m

/s

Scalar Product of Two Vectors

Scalar Product

Dot Products of Unit Vectors

Work - on and by

Mechanical Energy

Kinetic Energy

Kinetic Energy

Kinetic Energy

The Work-Energy Theorem

D









Hooke’s Law

Hooke’s Law

4.00 J

1 2

0.100 m

800 N m 1 2

4.00 J

12.0 J

W



m v



m v



m v



m v



m v

Work done by a general variable force:

Work done by a general variable force:









F

m a

F

m g



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