Download Chapter 7

86
Chapter 7 — Energy
invented to make everything alright. We feel it is important to show that the invention of other forms of
energy is not whimsical or a farce. A newly invented form of energy would be false if the new form was not
consistent with the conservation law and the other forms of energy.
Teaching Tips This section presents a subtle point that may be skipped if it doesn't match your course
goals. It is an attempt to address a common attitude of students that rules are made without rhyme or reason.
7-11 POWER
Goals
• Introduce the concept of power as the rate of converting energy from one form to another per unit time.
• Define a watt and a horsepower.
• (Computing) Calculate the power required to keep a car moving at constant speed.
• (Computing) Calculate the electrical energy converted by a motor.
• (Problem Solving) Calculate the power for a common event.
Content Power is the measure of the energy transformed divided by the time elapsed. Two common units
of power are the horsepower and the watt. A horsepower is about 750 watts.
Teaching Tips This is a short section that can be skipped if the material is not germane to your goals. If
you are into applications, this is a good section to use as a springboard for talking about local power needs.
Computing Power We calculate the power needed to overcome frictional effects and keep a car moving at a
constant speed. We also calculate the electrical energy used by a motor during a period of time. We show
that 1 kWh = 3.6 million joules. A question asks for the energy used by a light bulb in 8 hours.
Problem Solving 7.6 We calculate the power requirements of an accelerating automobile.
Computer Animations Active Figure Animations are available on the Multimedia Manager Instructor’s
Resource CD. They are organized by textbook chapter, and each animation comes within a shell that
provides information on how to use the animation, exploration activities, and a short quiz.
Answers to the Conceptual Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Momentum is a vector. The initial momentum of the system is zero.
Energy comes in many flavors of which kinetic is only one.
Momentum is always conserved in a collision whereas kinetic energy is rarely conserved.
Momentum and energy are entirely different quantities so it makes no sense to say one is bigger than
the other. This would be similar to saying that a person is taller than his weight.
Yes. If the kinetic energy is zero all objects in the system must be at rest.
No. Two identical cars approaching each other at the same speed have zero total momentum but they
do not have zero kinetic energy.
The motorboat and water skier because the supertanker is not moving and has zero kinetic energy.
The Chevy has four times the kinetic energy because the kinetic energy is proportional to the square of
the speed.
The minivan has the larger kinetic energy because the kinetic energy is proportional to the mass.
The sports car now has the larger kinetic energy because the kinetic energy is proportional to the square
Chapter 7 — Energy
87
of the speed and only the first power of the mass.
11. Their kinetic energies are the same because their speeds are the same.
12. Kinetic energy does not depend on direction and is therefore constant. The length of the momentum
vector is constant but, as the jet’s direction changes, so does the direction of its momentum vector.
13. Two balls will leave the other side with the same speed as the incoming balls. This conserves both
momentum and kinetic energy.
14. They will rebound with the same speeds, conserving kinetic energy and momentum.
15. Because work is the force in the direction of motion times the distance moved, we must know how high
she lifted the ball.
16. As Sally carries the bags to the truck, the force she applies is perpendicular to the displacement and
therefore does no work. She does the same work as Bill.
17. The kinetic energy decreases because the force and the displacement are in opposite directions.
18. The force, and therefore the resulting acceleration, is perpendicular to the velocity and so only the
direction of the velocity changes. The kinetic energy is constant but the direction of the momentum
changes.
19. The change in kinetic energy has to be the same because the same work is done in both cases. The
lighter sled will not be pushed for as many seconds, so the impulse is less on the lighter sled. The
lighter sled will have less final momentum.
20. The change in momentum has to be the same because the same impulse is imparted in both cases. The
lighter sled will be pushed farther and so more work is done on it than on the heavier sled. The lighter
sled will have more final kinetic energy.
21. Although the tractor does positive work on the trailer, the frictional forces and air resistance do an equal
amount of negative work so the kinetic energy (and speed) therefore remain constant.
22. At points A and B the gravitational force is doing positive work and the satellite’s kinetic energy is
increasing. At point C the gravitational force is doing negative work and the satellite’s kinetic energy
is decreasing.
23. Force A does less work because the component of the force in the direction of the displacement is
smaller.
24. Both forces do the same work because the component of the force in the direction of the displacement
is the same in each case.
25. The change in kinetic energy is equal to the work. Therefore, the same work must be performed on each
car to stop the car. Because the forces are equal, the distances must be equal.
26. The car with the larger mass will have the smaller kinetic energy and will therefore stop in the shorter
distance.
27. No, the kinetic energy lost by one object during a collision could be converted to many different forms
of energy, including deformation, sound, thermal energy, etc.
28. If you change an object's direction, you change its momentum, but if it still has the same speed, its
kinetic energy is unchanged. It is not possible to change its kinetic energy without also changing the
magnitude of its momentum.
29. 3 N × 3 m = 9 J and 4 N × 2 m = 8 J. Therefore, the first does the most work.
30. 5 N × 3 m = 15 J and 4 N × 4 m = 16 J. Therefore, the second does the most work.
31. Only differences in potential energy are important. The choice for the zero of potential energy is
arbitrary and could therefore be chosen as 2 meters above the floor.
32. It must be converted into thermal energy.
33. What is gained by the kinetic energy is lost by the gravitational potential energy, maintaining the
mechanical energy at a constant value.
88
Chapter 7 — Energy
34. If the ball is considered to be the system, only the mechanical energy is conserved. If Earth is added to
the system, momentum is also conserved.
35. The potential energy is a maximum at either end where the pendulum bob reaches its greatest height.
The kinetic energy is a maximum at the midpoint where the pendulum bob reaches its lowest height
and, consequently, its lowest potential energy.
36. The force of the string is always acting perpendicular to the velocity and does no work.
37. The friction causes the mechanical energy of the pendulum to decrease. Therefore the gravitational
potential energy is less at the end of each swing and the height decreases.
38. The loss in mechanical energy must be due to the work done by a force such as friction.
39. As the satellite moves closer to Earth, its gravitational potential energy decreases and its kinetic energy
increases. The reverse happens as the satellite moves farther from Earth.
40. The catapult's potential energy in its "loaded" position is transformed into kinetic energy of the
spaceship and catapult arm. Some of the ship's kinetic energy is converted into gravitational potential
energy as it leaves Earth and then back into kinetic energy as it approaches the Moon. Finally, this
kinetic energy is turned into heat and energy of distortion as the spaceship hits the Moon.
41. When the ball is at its maximum height, the basketball player pushes downward on it, doing work on it.
This work plus the ball's gravitational potential energy represents the total mechanical energy of the
ball. As the ball falls, it loses potential energy and gains kinetic energy. When it strikes the floor, it
distorts and its kinetic energy is transformed into elastic potential energy in its distortion (and that of
the floor to a lesser extent). When the ball and floor "spring" back, much of that elastic potential
energy returns to kinetic energy of the ball. The missing kinetic energy shows up as thermal energy in
the ball and floor (to be discussed in Chapter 13).
42. d) They all hit the ground with the same speed.
43. When the pucks approach each other, the magnetic repulsions do work on the pucks, slowing one down
and speeding the other up. During this transformation, kinetic energy is transformed into magnetic
potential energy, which is then returned to the pucks as kinetic energy. The total mechanical energy of
the system remains constant.
44. When the puck stretches the spring, it does work on it. This transforms the puck's kinetic energy into
elastic potential energy stored in the spring. This elastic potential energy is then transformed back into
kinetic energy. At all times, the total mechanical energy of the puck and spring stays constant if we
ignore the small amount that shows up as thermal energy in the spring. When the puck is suspended
from the ceiling, the mechanical energy stays the same but we must include gravitational potential
energy.
45. The work done in plowing through the dirt or sand reduces the kinetic energy (and hence the speed) of
the truck.
46. The soft sand absorbs the kinetic energy of the leg and does not return it like a springy track would.
47. The chemical potential energy is converted to electrical energy in the battery. This energy is converted
to thermal energy in the socks.
48. The brake shoes do work on the brake drums converting the car's kinetic energy to thermal energy.
49. We cannot completely recover the energy transformed due to frictional effects. (This is discussed more
fully in Chapter 14.)
50. The book loses 20 joules of energy due to the work done by friction as the book goes up the ramp. It
will lose another 20 joules in sliding down to end up with 160 joules of kinetic energy.
51. The rating tells you how much work can be done in one second, not the total amount that can be done.
If the winch will run for more than 1 second, it can do more than 600 joules of work.
52. If the battery is truly twice as powerful, it will be able to deliver twice as much energy in each second
making the bulb burn brighter. Since power only deals with rate of energy delivery, we have no way of
89
Chapter 7 — Energy
knowing how long it will last without knowing the initial chemical energy stored in the battery.
53. Kilowatt-hour is an energy unit because kilowatt is a power unit and multiplying power by a time gives
energy.
54. 1200 J ÷ 10 s = 120 watts, 5000 J ÷ 50 s = 100 watts. Valerie is more powerful.
55. Kilowatt-hour is an energy unit because kilowatt is a power unit and multiplying power by a time gives
energy.
56. Watt is a unit of power. The rest are energy units.
Answers to the Exercises
1.
2.
3.
(1800 kg ) ( 20 m s ) = 3.6 × 105 J
2
KE = 12 mv 2 = 12 ( 84 kg ) (10 m s ) = 4200 J
pi = ( 4 kg ) ( 6 m s ) − (1 kg ) ( 6 m s ) = 18 kg ⋅ m s
p f = ( 4 kg ) ( 3 m s ) + (1 kg ) (12 m s ) = 24 kg ⋅ m s
2
2
KEi = 12 ( 4 kg ) ( 6 m s ) + 12 (1 kg ) ( 6 m s ) = 90 J
2
2
KE f = 12 ( 4 kg ) ( 3 m s ) + 12 (1 kg ) (12 m s ) = 90 J
KE =
1
2
mv 2 =
2
1
2
momentum is not conserved
kinetic energy is conserved
Momentum is always conserved in a collision. This collision could not have taken place.
4.
pi = ( 4 kg ) ( 6 m s ) − (1 kg ) ( 6 m s ) = 18 kg ⋅ m s
p f = ( 4 kg ) ( 2 m s ) + (1 kg ) (10 m s ) = 18 kg ⋅ m s
KEi =
KE f =
( 4 kg ) ( 6 m s ) + 12 (1 kg ) ( 6 m s ) = 90 J
2
2
1
1
= 58 J
2 ( 4 kg ) ( 2 m s ) + 2 (1 kg ) (10 m s )
2
1
2
momentum is conserved
2
kinetic energy is not conserved
Momentum is always conserved in a collision. This collision could have taken place.
( 4 kg ) ( 5 m s ) = 50 J
2
KE f = 12 ( m1 + m2 ) v 2 = 12 ( 5 kg ) ( 4 m s ) = 40 J
⇒ ∆KE
2
2
KEi = 12 m1v12 + 12 m2 v22 = 12 ( 3 kg ) ( 6 m s ) + 12 ( 2 kg ) ( 4 m s ) =
2
KE f = 12 ( m1 + m2 ) v 2f = 12 ( 5 kg ) ( 2 m s ) = 10 J
⇒ ∆KE
KE f = KEi + Fd = 0 + ( 0.7 N ) ( 3 m ) = 2.1 J
Wby hand = 40 N ( 0.5 m ) = 20 J
5. KEi =
6.
7.
8.
1
2
m1v12 + 0 =
2
1
2
= − 10 J
70 J
= − 60 J
∆KE = Fnet ∆x = 0; Fnet = 0
KE f = KEi = 12 J
9.
10.
Fnet =
KE f − KEi
∆x
=
27 J − 5 J
= 11 N
2m
KE f = KEi + W = 12 J − ( 0.8 N ) ( 4 m ) = 8.8 J
11. The angle between the net force vector and the displacement vector is greater than 90° for every small
90
Chapter 7 — Energy
∆t over the next six months. The work done during this time is negative.
∆KE = 0 over a year's time. (Speed of Earth is the same each Jan 4th.) Work done during this time is
zero.
12. No work is performed because the gravitational force is perpendicular to the circular path.
2
13. W = mg ∆h = ( 55 kg ) 10 m s ( 720 m ) = 396 kJ
(
)
14. ∆GPE = mg ∆h = ( 65 kg ) (10 m s2 ) ( 3 m ) = 1950 J
(
15. a) mg ∆h = − ( 0.145 kg ) 10 m s
2
) (6 m )
= − 8.7 J
b) KE f = KEi + Work = 8.7 J − 8.7 J = 0
c) v f = 0
16. GPE = mgh = Wh = ( 50 N ) (1.5 m ) = 75 J assuming that the value for the gravitational potential
energy is chosen to be zero at the floor.
17. KE = mg ∆h = ( 0.5 kg ) (10 m s2 ) ( 6 m ) = 30 J
18. a) GPEf can be chosen to be zero; same
b) KE f = GPEi = mgh = ( 2 kg ) (10 m/s2 ) ( 4 m ) = 80 J; same
c) v =
2 × 80 J
= 8.94 m/s; same
2 kg
2 KE
=
m
d) p = mv = ( 2 kg )(8.94 m/s ) = 17.9 kg ⋅ m/s; The two blocks have velocities in different directions,
so the momenta must be different.
19. KE = mg ∆h = (1200 kg ) (10 m s2 ) (12 m ) = 144 kJ
20. a) GPE = mgh = ( 0.1 kg ) (10 m s 2 ) ( 5 m ) = 5 J
b) zero
c) KE f = GPEi + KEi = 5 J + 6 J = 11 J
d) KEi is independent of direction. The answer to part c) would not change.
( 300 lb ) ( 4 ft ) = 1500 ft ⋅ lb s ⎡ 1 hp ⎤ = 2.73 hp
∆E
W ∆h
=
=
21. P =
(
)⎢
⎥
∆t
∆t
0.8 s
⎣ 550 ft ⋅ lb s ⎦
∆E
22. P =
=
∆t
1
2
mv 2
=
∆t
1
2
(80 kg ) (10 m s )
3s
2
= 1330 W
23. ∆E = P∆t = (15 W ) (8 h ) = 120 Wh = 0.12 kWh
⎡ 1h ⎤
24. ∆E = P∆t = ( 0.8 kW ) ( 3 min ) ⎢
⎥ = 0.04 kWh
⎣ 60 min ⎦
Answers to the Problems in Problem Solving
1. KE =
1
2
mv 2 =
KE =
1
2
mv 2 =
2.
( 50 kg ) ( 4 m s ) = 400 J
2
1
30,000 kg ) ( 35 m s ) = 1.84 × 107 J
2 (
1
2
2
91
Chapter 7 — Energy
3. v =
2 KE
=
m
800,000 J
= 28.3 m s
1000 kg
4. v =
2 KE
=
m
1200 J
= 3.87 m s
80 kg
5.
pi = m1v1 + m2 v2 = (1 kg ) ( 8 m s ) + ( 2 kg ) ( −4 m s ) = 0
p f = m1v1 + m2 v2 = (1 kg ) ( −6 m s ) + ( 2 kg ) ( 3 m s ) = 0
KEi =
1
2
KE f =
m1v12 + 12 m2 v22 =
m1v12 +
1
2
1
2
m2 v22 =
(1 kg ) (8 m s ) + 12 ( 2 kg ) ( −4 m s ) = 48 J
2
2
1
1
= 27 J
2 (1 kg ) ( −6 m s ) + 2 ( 2 kg ) ( 3 m s )
2
1
2
2
Therefore, momentum is conserved, while kinetic energy is not conserved.
6.
pi = m1v1 + m2 v2 = (1 kg ) ( 6 m s ) + 0 = 6 kg ⋅ m s
p f = m1v1 + m2 v2 = (1 kg ) ( −2 m s ) + ( 2 kg ) ( 4 m s ) = 6 kg ⋅ m s
KEi =
1
2
KE f =
m1v12 + 12 m2 v22 =
m1v12 +
1
2
1
2
m2 v22 =
(1 kg ) ( 6 m s ) + 0 = 18 J
2
2
1
1
2 (1 kg ) ( −2 m s ) + 2 ( 2 kg ) ( 4 m s )
2
1
2
= 18 J
Therefore, both momentum and kinetic energy are conserved.
7.
⎛v ⎞
p f = ( m + 3 m ) ⎜ o ⎟ = mvo
⎝4⎠
pi = mvo
⇒
2
KEi =
KE f =
2
o
1
2
mv
1
2
v
( 4 m ) ⎛⎜ o ⎞⎟ =
⎝4⎠
1
2
pi = p f
⎡1⎤
mvo2 ⎢ ⎥
⎣4⎦
⎛ 3⎞
∆KE = − ⎜ ⎟ ( 12 mvo2 )
⎝4⎠
8. pi = mvo
mv 12
⎛ −v ⎞
⎛ 3v ⎞
pf = m ⎜ o ⎟ + 4 m ⎜ o ⎟ = − o +
mvo = mvo
5
10
⎝ 5 ⎠
⎝ 10 ⎠
KEi = 12 mvo2
2
2
⎛v ⎞
⎛ 3v ⎞
KE f = m ⎜ o ⎟ + 12 ( 4 m ) ⎜ o ⎟ =
⎝5⎠
⎝ 10 ⎠
2
1
∆KE = − 0.6 ⎡⎣ 2 mvo ⎤⎦
1
2
9.
36 ⎤
⎡1
mvo ⎢ +
=
⎣ 25 100 ⎦⎥
pi = m1v1 + m2 v2 = (1 kg ) ( 6 m s ) + 0 = 6 kg ⋅ m s = p f
vf =
pf
m1 + m2
KE f =
10.
1
2
pi = p f
1
2
=
6 kg ⋅ m s
= 2m s
1 kg + 2 kg
( m1 + m2 ) v 2f = 12 (1 kg + 2 kg ) ( 2 m s )
= ( 0.01 kg ) ( 900 m s ) = 9 kg ⋅ m s
2
= 6J
⇒
1
2
pi = p f
mvo ( 0.4 )
92
Chapter 7 — Energy
vf =
pf
m
KEi =
KE f =
=
9 kg ⋅ m s
= 4.48 m s
2.01 kg
( 0.01 kg ) ( 900 m s ) = 4050 J
2
1
= 20.2 J
2 ( 2.01 kg ) ( 4.48 m s )
2
1
2
4050 − 20.2
= 0.995
4050
fraction lost =
− 1 ( 0.5 kg ) ( 40 m s )
∆KE
− 1 mv 2
11. F =
= 2
= 2
= − 2000 N
0.20 m
d
d
where the minus sign indicates that the force is opposite the motion.
2
12.
∆KE
F =
=
d
1
2
mv 2
=
d
1
2
(80 kg ) ( 9 m s )
10 m
2
= 324 N
13. KE f = KEi + Fd = 0.4 J + ( 0.4 N ) ( 0.2 m ) = 0.48 J
14.
KE f = KEi + Fd = 0.2 J + ( 0.2 N ) ( 0.5 m ) = 0.3 J
15. Using the data for 70 mph, we have
103 ft 202 ft
d tot = 12 d react + 14 d brake =
+
= 107 ft
2
4
16. Using the data for 90 mph, we have
132 ft 363 ft
d tot = 12 d react + 14 d brake =
+
= 157 ft
2
4
17. W = Fd = (100 N ) ( 3 m ) = 300 J
18. W = Fd = (120 N ) (1.2 m ) = 144 J
19. In the second collision, block B will have half the speed, and hence one-fourth the kinetic energy, after
the collision. This kinetic energy is transformed into potential energy as block B moves up the ramp.
Block B (and block A) moves only 10 cm up the ramp.
2
20. KE f = PEi = mg ∆h = (10 kg ) 9.8 m s ( 0.8 m ) = 78.4 J
(
)
KE f = KEi + Fd = 78.4 J − ( 25 N ) ( 2 m ) = 28.4 J
vf =
2 ( KE f
)
= 2.38 m s
m
21. ∆GPE = mg ∆h = ( 80 kg ) ( 9.8 m s2 ) ( −10 m ) = − 7840 J
22. a) GPEcounter = Wh = ( 2 N ) (1 m ) = 2 J
GPE shelf = Wh = ( 2 N ) ( 2 m ) = 4 J
b) 2 J
c) The answers to part a) are 2 J lower; the answer in b) does not change.
2
m
m
⎛m⎞
23. [ KE ] = kg ⎜ ⎟ ;
[ work ] = N ⋅ m = kg 2 m; [GPE ] = kg 2 m
s
s
⎝ s ⎠
93
Chapter 7 — Energy
24. KE =
1
2
mv 2 =
1
2
m2v 2
p2
=
2m
m
(
25. KE f = KEi − ∆GPE = 0 − mg ∆h = − ( 30 kg ) 9.8 m s
(
26. ∆KE = − ∆GPE = − mg ∆h = − ( 85 kg ) 9.8 m s
27. v =
2 KE
=
m
−2 ∆ GPE
=
m
28. v =
2 KE
=
m
2 × 588 J
= 6.26 m s
30 kg
2 KE
−2 ∆ GPE
=
= −2 g ∆h =
m
m
30. At the lowest point:
vbob = 3.13 m s
( from prob 29 )
= 3.27 m s2
T = ( 2 kg ) 3.27 m s + 9.8 m s
31. h =
1
2
2
mv 2
=
W
1
2
) ( −10 m ) = 8330 J
− 2 ( 9.8 m s2 ) ( −0.5 m ) = 3.13 m s
2
ac =
(
) ( − 2 m ) = 588 J
− 2 ( 9.8 m s2 ) ( −9.8 m ) = 13.9 m s
−2 g ∆h =
29. v =
2
( 3.13 m s )
vbob
=
3m
R
T − mg = mac
2
2
2
)
up
= 26.1 N
( 40 m s ) = 81.6 m
mv 2
v2
=
=
2
mg
2g
2 × 9.8 m s
2
32. a) W = Fnet d = ( 5 N − 1 N ) ( 82 m ) = 328 J
1 N ) ( 80 m s ) 2
(
Wv 2
mv =
=
= 327 J
2
2g
2 × 9.8 m s
c) The change in kinetic energy should equal the work done by the net force. (The small difference is
due to round off.)
−∆KE
327 J
=
= 327 m
d) ∆h =
mg
1N
b) KE =
33.
1
2
∆KE
P =
=
∆t
2
1
2
mv 2
=
∆t
1
2
(80 kg ) ( 9 m s )
2s
2
= 1620 W
1
1
mv 2
( 2000 kg ) ( 30 m s ) = 2.25 × 105 W
∆KE
= 2
= 2
34. P =
∆t
∆t
4s
35. The work is converted to gravitational potential energy as the bucket is raised.
2
36. The work is transformed into heat by the frictional force.
J
⎛m⎞
37. N ⎜ ⎟ =
s
⎝s⎠
P
12,300 W
=
= 549 N
38. F =
v
22.4 m s