Transcript Work and Energy - Dripping Springs ISD

Work and Energy
The Ninja, a roller coaster at Six Flags over
Georgia, has a height of 122 ft and a speed of
52 mi/h. The potential energy due to its
height changes into kinetic energy of motion.
Energy
Energy is anything that can be converted into work; i.e., anything that can
exert a force through a distance.
Energy is the capability for doing work.
Definition of Work
Work is a scalar quantity equal to the
product of the displacement x and the
component of the force Fx in the
direction of the displacement.
Work = Force component X displacement
Work = Fx d
Positive Work
F
x
Force F contributes to displacement x.
Example: If F = 40 N and x = 4 m, then
Work = (40 N)(4 m) = 160 Nm
Work = 160 J
1 Nm = 1 Joule (J)
Negative Work
x
f
The friction force f opposes the displacement.
Example: If f = -10 N and x = 4 m, then
Work = (-10 N)(4 m) = - 40 J
Work = - 40 J
Resultant Work or Net Work
Resultant work is the algebraic sum of
the individual works of each force.
x
f
F
Example: F = 40 N, f = -10 N and x = 4 m
Work = (40 N)(4 m) + (-10 N)(4 m)
Work = 120 J
Resultant Work (Cont.)
Resultant work is also equal to the
work of the RESULTANT force.
4 m -10 N
Example: Work = (F - f) d
Work = (40 - 10 N)(4 m)
Work = 120 J
40 N
Work of a Force at an Angle
Work = Fx x
F = 70 N
d = 12 m
60o
Work = (F cos ) d
Work = (70 N) Cos 600 (12 m) = 420 J
Work = 420 J
Only the x-component of
the force does work!
Example: A lawn mower is pushed a horizontal
distance of 20 m by a force of 200 N directed at
an angle of 300 with the ground. What is the
work done by this force?
F
d = 20 m
300
F = 200 N
Work = (F cos  ) d
Work = (200 N)(20 m) Cos 300
Work = 3460 J
Note: Work is
positive since Fx
and x are in the
same direction.
Example: A 40-N force pulls a 4-kg block a
horizontal distance of 8 m. The rope makes an angle
of 350 with the floor and uk = 0.2. What is the work
done by each force acting on the block, and what is
the net work?
P
x

1. Draw sketch and
find given values.
P = 40 N; x = 8 m, uk = 0.2;  = 350; m = 4 kg
2. Draw free-body
diagram showing
all forces. (Cont.)
Work = (F cos ) d
fk
mg
x
n
P
40 N
350
+d
8m
Example (Cont.): Find Work Done by Each Force.
fk
n
W = mg
x
P
40 N
350
+d
8m
P = 40 N; x = 8 m, uk = 0.2;
 = 350; m = 4 kg
4. First find work of P.
Work = (P cos ) d
WorkP = (40 N) cos 350 (8 m) = 262 J
Example (Cont.):
fk
n
W = mg
d
P
40 N
350
+d
P = 40 N; x = 8 m, uk = 0.2;
 = 350; m = 4 kg
WorkP = 262 J
8m
5. Next find work of friction. Recall: fk = mk n
n = (4 kg)(9.8 m/s2) – (40 N)sin 350 = 16.3 N
fk = mk n = (0.2)(16.3 N);
fk = 3.25 N
Example (Cont.):
WorkP = 262 J
6. Work of friction (Cont.)
fk = 3.25 N; d = 8 m
fk
n
W = mg
d
P
40 N
350
+d
8m
Workf = (-3.25 N)(8 m) = -26.0 J
7. The resultant work is the sum of all works:
262 J – 26 J
(Work)R = 236 J
Work and Kinetic Energy
A resultant force changes the velocity of an
object and does work on that object.
vo
m
vf
x
F
F
m
a
Work  Fx  ( ma ) x;
Work  mv  mv
1
2
2
f
1
2
v v
2
0
2
f
2
0
2x
The Work-Energy Theorem
Work is equal
to the change
in ½mv2
If we define kinetic energy as ½mv2 then we
can state a very important physical principle:
The Work-Energy Theorem: The work
done by a resultant force is equal to the
change in kinetic energy that it produces.
Example: A 20-g projectile strikes a mud
bank, penetrating a distance of 6 cm before
stopping. Find the stopping force F if the
6 cm
entrance velocity is0 80 m/s.
80 m/s
Work = ½
mvf2 -
½ mvo
x
2
F=?
F x = - ½ mvo2
F (0.06 m) = - ½ (0.02 kg)(80 m/s)2
F (0.06 m)= -64 J
F = -1067 N
Work to stop bullet = change in K.E. for bullet
Example: A bus slams on brakes to avoid an
accident. The tread marks of the tires are 80 m
long. If mk = 0.7, what was the speed before
Work = DK
applying brakes?
Work = F(cos ) x
f = mk.n = mk mg
Work = - mk mg x
2 = -m mg x
-½DKmv
=o Work k
25 m
f
0
DK = ½ mvf2 - ½ mvo2
vo =
vo = 2(0.7)(9.8 m/s2)(80 m)
2mkgx
vo = 33.1 m/s
Power
Power is defined as the rate at which
work is done: (P = W/t )
F
E
t
m
4s
P 
t
10 kg
h
20 m
mg
mgh (10 )( 9.8)( 20 )
P

t
4
P  490 J/s or 490 watts (W)
Units of Power
One watt (W) is work done at
the rate of one joule per second.
1 W = 1 J/s
Example of Power
What power is consumed in lifting
a 70-kg robber 1.6 m in 0.50 s?
Fh mgh
P

t
t
2
(70 kg)(9.8 m/s )(1.6 m)
P
0.50 s
Power Consumed: P = 2195 W
Example : A 100-kg cheetah moves from rest
to 30 m/s in 4 s. What is the power?
Recognize that work is equal to
the change in kinetic energy:
Work  mv  mv
P
1
2
2
f
mv 2f
(100 kg)(30 m/s) 2

4s
t
1
2
2
0
Work
P
m = 100 kg
t
1
2
1
2
Power Consumed: P = 11250 W
Power and Velocity
Recall that average or constant velocity is
distance covered per unit of time v = x/t.
P=
Fd
t
=F
d
t
P  Fv
Example : What power is
required to lift a 900-kg
elevator at a constant speed
of 4 m/s?
P = F v = mg v
P = (900 kg)(9.8 m/s2)(4 m/s)
P = 35280 W
v = 4 m/s
Summary
Potential Energy: Ability to do work
U

mgh
by virtue of position or condition.
Kinetic Energy: Ability to do work by
2
1
K

mv
2
virtue of motion. (Mass with velocity)
The Work-Energy Theorem: The work done by
a resultant force is equal to the change in
kinetic energy that it produces.
Work = ½ mvf2 - ½ mvi2
Summary (Cont.)
Power is defined as the rate at which P  Work
t
work is done: (P = W/t )
Work F r
Power 

time
t
P= F v
Power of 1 W is work done at rate of 1 J/s
CONCLUSION:
Work and Energy