OpenStax College Physics
Instructor Solutions Manual
Chapter 7
CHAPTER 7: WORK, ENERGY, AND ENERGY
RESOURCES
7.1 WORK: THE SCIENTIFIC DEFINITION
1.
How much work does a supermarket checkout attendant do on a can of soup he
pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and
kilocalories.
Solution
W  Fdcosθ  (5.00 N) (0.600 m) cos0  3.00 J
W  3.00 J 
1 kcal
 7.17  10 4 kcal
4186 J
2.
A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to
accomplish this task.
Solution
W  mgh  (75.0 kg) (9.80 m/s 2 ) (2.50 m)  1838 J  1.84  10 3 J
3.
(a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at
constant speed, assuming friction averages 100 N. (b) What is the work done on the
lift by the gravitational force in this process? (c) What is the total work done on the
lift?
Solution (a) Wc  Fdcosθ  (mg  Ff )  d


 (1500 kg) (9.80 m/s 2 )  100 N  40.0 m  5.92  10 5 J
(b) Wg  mgh  (1500 kg) (9.80 m/s 2 ) (40.0 m)   5.88 105 J
(c) The net force is zero, since the elevator moves at a constant speed. Therefore, the
total work done is 0 J .
4.
Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline.
Only 30% of the gasoline goes into useful work by the force that keeps the car moving
at constant speed despite friction. (See Table 7.1 for the energy content of gasoline.)
(a) What is the force exerted to keep the car moving at constant speed? (b) If the
required force is directly proportional to speed, how many gallons will be used to drive
108 km at a speed of 28.0 m/s?
OpenStax College Physics
Instructor Solutions Manual
Chapter 7
Solution (a) Wf  (2.0 gal) (1.2 108 J/gal) (0.30)  7.2 10 7 J
Wf  Ff d  Ff 
(b)
5.
Solution
Wf
7.2 10 7 J

 666.7 N  6.7 10 2 N
d 1.08 10 5 m
28.0 m/s
 2.0 gal  1.867 gal  1.9 gal
30.0 m/s
Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a
ramp that makes an angle of 20.0 with the horizontal. (See Figure 7.35.) He exerts a
force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be
certain to include the work he does on the crate and on his body to get up the ramp.

W  Fd  (mgsin θ  500 N)  d


 (85.0 kg) (9.80 m/s 2 ) (sin20 )  500 N  4.00 m  3.14  10 3 J
6.
How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in
Figure 7.36? Assume no friction acts on the wagon.
Solution
W  Fdcosθ  50.0 N  30.0 m  cos30  1.299  10 3 J  1.30  10 3 J
7.
A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a
35.0 N frictional force. He pushes in a direction 25.0 below the horizontal. (a) What is
the work done on the cart by friction? (b) What is the work done on the cart by the
gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the
force the shopper exerts, using energy
considerations. (e) What is the total work done
on the cart?
Solution (a) Wf  Fdcosθ  35.0 N  20.0 m  cos180   700 J
(b) Wg  Fdcosθ  35.0 N  20.0 m  cos90  0 J
(c) net W  Ws  Wf  0, or Ws  700 J
(d) Ws  Fdcosθ where   25, so that:
Ws
700 J
F

 38.62 N  38.6 N
d cosθ 20.0 m  cos25
(e) net W  Wf  Ws  700 J  700 J  0 J
8.
Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg,
down a 60.0 slope at constant speed, as shown in Figure 7.37 The coefficient of
friction between the sled and the snow is 0.100. (a) How much work is done by friction
as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on
the sled in this distance? (c) What is the work done by the gravitational force on the
sled? (d) What is the total work done?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 7
T
N
Ff
 = 60°
(a) Wf  Ff d cos θ; Ff  μs N  μs mg cos φ,
Wf  μs mg cos φ (d cos θ)
 (0.100) (90.0 kg) (9.80 m/s 2 ) (cos60.0 ) (30.0 m) (cos180  )
Wf  1.323  10 3 J   1320 J
(b) Wr  Td cos ; T  Ff  mg sin 
T  mg sin    s mg cos   mg (sin    s cos  )
Wr  mg (sin    s cos  )d cos
Wr  (90.0 kg) (9.80 m/s 2 ) [ sin60.0   (0.100) cos60.0 ] (30.0 m) (cos180 )
 2.1592 10 4 J   2.16 10 4 J
(c) Wg  Fg d cos   mg sin  (d cos  )
Wg  (90.0 kg) (9.80 m/s 2 ) (sin60.0 ) (30.0 m) (cos0)
 2.2915  10 4 J  2.29  10 4 J
(d) net W  Wf  Wr  Wg  1.323 103 J  2.1592 10 4 J  2.2915 10 4 J  0 J
(Since the sled moves at constant speed, this must be so.)
7.2 KINETIC ENERGY AND THE WORK-ENERGY THEOREM
9.
Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an
80.0-kg astronaut in orbit moving at 27,500 km/h.
Solution
27500 km/h 
1000 m
1h

 7.638  10 3 m/s
1 km
3600 s
1000 m
1h
110 km/h 

 30.56 m/s
1 km
3600 s
1
KE tr  mv 2  0.5 (20,000 kg) (30.56 m/s) 2  9.336  10 6 J  9.34  10 6 J
2
1 2
KE as  mv  0.5 (80.0 kg) (7.638  10 3 m/s) 2  2.334  10 9 J  2.33  10 9 J
2
KE tr 9.34  10 6 J
1


9
KE as 2.33  10 J 250
OpenStax College Physics
Instructor Solutions Manual
Chapter 7
10.
(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a
65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for
the movement of larger animals would relate to metabolic rates.
Solution
(a) KE 
1
1
1
mv 2 , so for the same KE: melephantv 2 elephant  msprinterv 2 sprinter , and
2
2
2
velephant 
m
sprinter
melephant
v 2 sprinter 
65.0 kg
(10.0 m/s) 2  1.47 m/s
3000 kg
(b) If the elephant and the sprinter accelerate to a final velocity of 10.0 m/s, then the
elephant would have a much larger kinetic energy than the sprinter. Therefore,
the elephant clearly has burned more energy and requires a faster metabolic
output to sustain that speed.
11.
Confirm the value given for the kinetic energy of an aircraft carrier in Table 7.1. You
will need to look up the definition of a nautical mile (1 knot = 1 nautical mile/h).
Solution
2240 lb 0.4539 kg

 9.1506  10 7 kg;
1 ton
1 lb
1.1516 mi/h 1609 m
1h
30 knot 


 15.44 m/s
1 knot
1 mi
3600 s
1
KE  mv 2  0.5 (9.1506  10 7 kg) (15.44 m/s) 2  1.091  1010 J  1.1  1010 J
2
(to two significant figures)
12.
(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h
in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose
instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00
m. Calculate the force exerted on the car and compare it with the force found in part
(a).
Solution
(a) 90.0 km/h 
90,000 ton 
1000 m
1h

 25.0 m/s
1 km
3600 s
1
1
2
W  Fd  mv 2  mv0 
2
2
 mv0
 (950 kg) (25.0 m/s) 2

 2.474  10 3 N   2470 N
2d
2  120 m
2
 mv0
 (950 kg) (25.0 m/s) 2

 1.484  10 5 N   1.48  10 5 N
(b) F 
2d
2  2.00 m
Fb
 60.0
Fa
2
F
OpenStax College Physics
Instructor Solutions Manual
Chapter 7
13.
A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an
immovable object without damage to the body of the car. The bumper cushions the
shock by absorbing the force over a distance. Calculate the magnitude of the average
force on a bumper that collapses 0.200 m while bringing a 900-kg car to rest from an
initial speed of 1.1 m/s.
Solution
Using the work energy theorem, net W 
14.
Boxing gloves are padded to lessen the force of a blow. (a) Calculate the force exerted
by a boxing glove on an opponent’s face, if the glove and face compress 7.50 cm
during a blow in which the 7.00-kg arm and glove are brought to rest from an initial
speed of 10.0 m/s. (b) Calculate the force exerted by an identical blow in the gory old
days when no gloves were used and the knuckles and face would compress only 2.00
cm. (c) Discuss the magnitude of the force with glove on. Does it seem high enough to
cause damage even though it is lower than the force with no glove?
Solution
(a) net W  Fd 
15.
Using energy considerations, calculate the average force a 60.0-kg sprinter exerts
backward on the track to accelerate from 2.00 to 8.00 m/s in a distance of 25.0 m, if he
encounters a headwind that exerts an average force of 30.0 N against him.
1
1
2
mv 2  mv0  Fdcosθ ,
2
2
2
2
2
mv  mv0
(900 kg) (0 m/s)  (900 kg) (1.12 m/s) 2
F

  2.8  10 3 N
2dcosθ
2 (0.200 m) cos0
The force is negative because the car is decelerating.
1 2 1
2
mv  mv0 , since v  0 ,
2
2
2
2
 mv0
mv0
on the face, or
F
on the glove  F 
2d
2d
(7.00 kg)(10.0 m/s) 2
F
 4.67  10 3 N
2(0.0750 m)
mv 2 (7.00 kg)(10.0 m/s) 2
(b) F 

 1.75  10 4 N
2d
2(0.0200 m)
(c) The force with the glove on is as if a 477 kg (1050 lb) mass were placed on the
person’s face. That definitely could do damage, even though it is less than the
effective 1790 kg mass of the force with no gloves.
Fw
Solution
F
net F
OpenStax College Physics
Instructor Solutions Manual
Chapter 7
1 2 1
1
2
2
mv  mv0  m (v 2  v0 )
2
2
2
m 2
60.0 kg
2
net F 
(v  v 0 ) 
(8.00 m/s) 2  (2.00 m/s) 2  72.0 N
2d
2  25.0 m
net F  F  FW  F  net F  FW  72.0 N  30.0 N  102 N
net F  d 


7.3 GRAVITATIONAL POTENTIAL ENERGY
16.
A hydroelectric power facility (see Figure 7.38) converts the gravitational potential
energy of water behind a dam to electric energy. (a) What is the gravitational
potential energy relative to the generators of a lake of volume 50.0 km3 (
mass  5.00  1013 kg ), given that the lake has an average height of 40.0 m above the
generators? (b) Compare this with the energy stored in a 9-megaton fusion bomb.
Solution (a) ΔPE g  mgh, so that ΔPE g  (5.00 1013 kg) (9.80 m/s 2 ) (40.0 m)  1.96 1016 J
(b)
17.
Elake 1.96  1016 J

 0.52. The energy stored in the lake is approximately half
E bomb
3.8  1016 J
that of a 9-megaton fusion bomb.
(a) How much gravitational potential energy (relative to the ground on which it is
built) is stored in the Great Pyramid of Cheops, given that its mass is about 7 10 9 kg
and its center of mass is 36.5 m above the surrounding ground? (b) How does this
energy compare with the daily food intake of a person?

Solution (a) PE  mgh  (7  10 9 kg) (9.80 m/s 2 ) (36.5 m)  2.5  1012 J  3  1012 J
Epyr 2.5 1012 J

 2.087 105  2 105
(b)
7
Efood 1.2 10 J
18.
Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises
it 2.5 m from the ground to a branch. (a) How much work did the bird do on the
snake? (b) How much work did it do to raise its own center of mass to the branch?
Solution (a) PE  mgh  (0.75 kg) (9.80 m/s 2 ) (2.5 m)  1.8375 J  1.8 J
(b) PE  mgh  (0.350 kg) (9.80 m/s 2 ) (2.5 m)  8.575 J  8.6 J
19.

In Example 7.7, we found that the speed of a roller coaster that had descended 20.0 m
was only slightly greater when it had an initial speed of 5.00 m/s than when it started
from rest. This implies that PE >> KEi . Confirm this statement by taking the ratio of
PE to KEi . (Note that mass cancels.)


OpenStax College Physics
Instructor Solutions Manual

Chapter 7

Solution
ΔPE
mgh
2 gh 2 9.80 m/s 2 20.0 m 



 15.7
KEi (1/ 2)mvi2
vi2
(5.00 m/s) 2
Thus, the change in potential energy is almost sixteen times larger than the initial
kinetic energy of the roller coaster.
20.
A 100-g toy car is propelled by a compressed spring that starts it moving. The car
follows the curved track in Figure 7.39. Show that the final speed of the toy car is
0.687 m/s if its initial speed is 2.00 m/s and it coasts up the frictionless slope, gaining
0.180 m in altitude.
Solution
1
1
2
2
2
2
mv0  mgh  mvf  v0  2 gh  vf
2
2

vf  (v0  2 gh)1 / 2  2.00m/s   2(9.80m/s 2 ) (0.180m)
2
2

1/2
 0.687m/s
21.
In a downhill ski race, surprisingly, little advantage is gained by getting a running
start. (This is because the initial kinetic energy is small compared with the gain in
gravitational potential energy on even small hills.) To demonstrate this, find the final
speed and the time taken for a skier who skies 70.0 m along a 30 slope neglecting
friction: (a) Starting from rest. (b) Starting with an initial speed of 2.50 m/s. (c) Does
the answer surprise you? Discuss why it is still advantageous to get a running start in
very competitive events.

Solution
1
1
2
2
2
2
mv0  mgh  mvf or v0  2 gh  vf where h  (70.0 m)sin30   35.0 m
2
2
v

0
m/s,
so
that
(a) 0
vf  2 gh  v02  2gh  2 (9.80 m/s 2 ) (35.0 m)  26.19 m/s  26.2 m/s
1
Using L  v0t  at 2 , where v0  0 m/s and a  g sin  ,
2
2L
2(70.0 m)
t

 5.35 s
g sin 
(9.80 m/s 2 )sin30 
(b) v0  2.50 m/s so that
vf  2 (9.80 m/s 2 ) (35.0 m)  (2.50 m/s) 2  26.31 m/s  26.3 m/s
Using v  v0  at , where v0  2.50 m/s and a  g sin  ,
v  v0
(26.31 m/s )  (2.50 m/s )
t

 4.86 s
g sin 
(9.80 m/s 2 ) sin 30
(c) The answer should not surprise you, since the initial velocity is small compared to
the final velocity. It is still advantageous to get a running start in very competitive
events because a half second could mean the difference between first and second
place.
OpenStax College Physics
Instructor Solutions Manual
Chapter 7
7.4 CONSERVATIVE FORCES AND POTENTIAL ENERGY
22.
A 5.00  105 - kg subway train is brought to a stop from a speed of 0.500 m/s in 0.400
m by a large spring bumper at the end of its track. What is the force constant k of the
spring?
Solution
1 2 1 2
kx  mv 
2
2
mv 2 (5.00  10 5 kg) (0.500m/s) 2
k 2 
 7.812  10 5 N/m  7.81  10 5 N/m
2
x
(0.400m)
23.
A pogo stick has a spring with a force constant of 2.50 10 4 N/m, which can be
compressed 12.0 cm. To what maximum height can a child jump on the stick using
only the energy in the spring, if the child and stick have a total mass of 40.0 kg?
Explicitly show how you follow the steps
in the Problem-Solving Strategies for Energy.
Solution
1 2
kx 2
2.50  10 4 N/m 0.120 m 
kx  mgh so that h 

 0.459 m
2
2mg
2(40.0 kg)(9.80 m/s 2 )


2
7.5 NONCONSERVATIVE FORCES
24.
A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-m-high rise as shown
in Figure 7.40. Find her final speed at the top, given that the coefficient of friction
between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the
incline assuming a straight-line path as shown in the figure.)
Solution
KE i Wnc  KE f  PE g
1
1
2
2
mv0  fd  mvf  mgh
2
2
1
h
1
2
2
mv0  mg cos  
 mvf  mgh
2
sin  2
2
2
v0  v f  2 gh (1   cot  )

v f  v  2 gh (1   cot  )
25.
2
0

1/ 2
 12.0 m/s 2  2 (9.80 m/s 2 ) (2.50 m) 


 (1  0.0800cot35)

1/2
 9.46 m/s
(a) How high a hill can a car coast up (engine disengaged) if work done by friction is
negligible and its initial speed is 110 km/h? (b) If, in actuality, a 750-kg car with an
initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its
starting point, how much thermal energy was generated by friction? (c) What is the
average force of friction if the hill has a slope 2.5 above the horizontal?

OpenStax College Physics
Instructor Solutions Manual
Chapter 7
Solution (a) Initially the car’s energy is all kinetic energy; finally it is all potential energy.
1000 m
1h
v0  110 km/h 

 30.56 m/s
1 km
3600 s
v2
1 2
(30.56 m/s) 2
mv0  mgh  h  0 
 47.63 m  47.6 m
2
2 g 2 (9.80 m/s 2 )
(b) Since the car coasts only to 22.0 m, some of the energy Ef must be lost to
thermal energy due to friction.
 v2

1 2
mv0  mgh  Ef  Ef  m  0  gh 
2
2

 (30.56 m/s) 2

 750 kg 
 (9.80 m/s 2 ) (22.0 m)   1.89  105 J
2


(c)
L
h

h
h
W sin 
, so that W  FL  F 
F
sin 
sin 
h
5
1.89  10 J sin2.5 
F
 375 N
22.0 m
h  L sin   L 


7.6 CONSERVATION OF ENERGY
26.
Using values from Table 7.1, how many DNA molecules could be broken by the energy
carried by a single electron in the beam of an old-fashioned TV tube? (These electrons
were not dangerous in themselves, but they did create dangerous x rays. Later model
tube TVs had shielding that absorbed x rays before they escaped and exposed
viewers.)
Solution
Eelec 4.0 10 15 J
n

 4 10 4 molecules
19
EDNA
10 J
27.
Using energy considerations and assuming negligible air resistance, show that a rock
thrown from a bridge 20.0 m above water with an initial speed of 15.0 m/s strikes the
water with a speed of 24.8 m/s independent of the direction thrown.
Solution
1
1
2
mvf2  mv0  mgh f by conservation of energy. So:
2
2

vf  (v0  2 gh)1 / 2  (15.0 m/s) 2  2 (9.80 m/s 2 ) (20.0 m)
2

1/ 2
 24.84 m/s  24.8 m/s
OpenStax College Physics
28.
Solution
29.
Solution
Instructor Solutions Manual
Chapter 7
If the energy in fusion bombs were used to supply the energy needs of the world, how
many of the 9-megaton variety would be needed for a year’s supply of energy (using
data from Table 7.1)? This is not as far-fetched as it may sound—there are thousands
of nuclear bombs, and their energy can be trapped in underground explosions and
converted to electricity, as natural geothermal energy is.
n
EWorld EnergyUse
Ebomb

4 10 20 J
 110 4 bombs
16
3.8 10 J
(a) Use of hydrogen fusion to supply energy is a dream that may be realized in the
next century. Fusion would be a relatively clean and almost limitless supply of energy,
as can be seen from Table 7.1. To illustrate this, calculate how many years the present
energy needs of the world could be supplied by one millionth of the oceans’ hydrogen
fusion energy. (b) How does this time compare with historically significant events,
such as the duration of stable economic systems?
(a) n 
Eoc / 10 6
E World Energy Use

(10 34 J) (10 -6 )
 2.5 10 7 years  25 million years!
4 10 20 J/yr
(b) This is much, much longer than human time scales.
7.7 POWER
30.
The Crab Nebula (see Figure 7.41) pulsar is the remnant of a supernova that occurred
in A.D. 1054. Using data from Table 7.3, calculate the approximate factor by which the
power output of this astronomical object has declined since its explosion.
Solution
P
10 28 W

 2  10 10 W. This power today is 1010 orders of magnitude smaller
37
P0 5  10 W
than it was at the time of the explosion.
31.
Suppose a star 1000 times brighter than our Sun (that is, emitting 1000 times the
power) suddenly goes supernova. Using data from Table 7.3: (a) By what factor does
its power output increase? (b) How many times brighter than our entire Milky Way
galaxy is the supernova? (c) Based on your answers, discuss whether it should be
possible to observe supernovas in distant galaxies. Note that there are on the order of
1011 observable galaxies, the average brightness of which is somewhat less than our
own galaxy.
Solution
P
5  10 37 W

 1.25  10 8  1  10 8
26
P0 1000 (4  10 W)
(b) Its power increases by about 5 times.
(c) Because a supernova is 5 times brighter than our galaxy, a supernova would be
(a)
OpenStax College Physics
Instructor Solutions Manual
Chapter 7
approximately 5 times brighter than the average brightness of a distant galaxy.
Therefore, it would seem reasonable that we might be able to observe
supernovas in distant galaxies by comparing their brightness today to earlier.
32.
A person in good physical condition can put out 100 W of useful power for several
hours at a stretch, perhaps by pedaling a mechanism that drives an electric generator.
Neglecting any problems of generator efficiency and practical considerations such as
resting time: (a) How many people would it take to run a 4.00-kW electric clothes
dryer? (b) How many people would it take to replace a large electric power plant that
generates 800 MW?
Solution
(a) n  4.00  10 3 W 
1 person
 40 people
100 W
1 person
(b) n  8.00  108 W
 8 million people
100 W


33.
What is the cost of operating a 3.00-W electric clock for a year if the cost of electricity
is $0.0900 per kW h ?
Solution
1 kW
365 24 h


 26.28 kW  h
3
1  10 W 1 y 1 d
$0.0900
cost  26.28 kW  h 
 $2.37
kW  h
34.
A large household air conditioner may consume 15.0 kW of power. What is the cost of
operating this air conditioner 3.00 h per day for 30.0 d if the cost of electricity is
$0.110 per kW h ?
Solution
3h
 1.35 103 kW  h
1d
$0.110
cost  1.35 103 kW  h  
 $148.50  $149
kW  h
35.
(a) What is the average power consumption in watts of an appliance that uses
5.00 kW  h of energy per day? (b) How many joules of energy does this appliance
consume in a year?
Solution
(a) P 
E  Pt  3.00 W 
E  Pt  15.0 kW  30 d 


E
5.00  10 3 J/s 3600 s 
 5.00 kW  h/d 
24 h 3600 s/h 
t
1.80  10 7 J

 208.33 W  208 W
8.64  10 4 s
24 h 3600 s

 6.57  10 9 J
(b) E  Pt  208.33 W  365 d 
1d
1h
OpenStax College Physics
36.
Solution
Instructor Solutions Manual
(a) What is the average useful power output of a person who does 6.00  10 6 J of
useful work in 8.00 h? (b) Working at this rate, how long will it take this person to lift
2000 kg of bricks 1.50 m to a platform? (Work done to lift his body can be omitted
because it is not considered useful output here.)


W
6.00 106 J

 208.3 J/s  208 W
t 8.00 h 3600 s/h 
W mgh


(b) W  mgh; P 
t
t
mgh 2000 kg  9.80 m/s 2 1.50 m 
t

 141.1 s  141 s
P
208.3 W
(a) P 

37.
Chapter 7

A 500-kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a
quarter of a mile) and encounters an average frictional force of 1200 N. What is its
average power output in watts and horsepower if this takes 7.30 s?
Solution Using the equation KEi  PEi  Wnc  OEi  KEf  PE f  OEf to express conservation
of energy and identifying that KEi  PEi  PEf  0 we see that ΔOE  KEf  Wnc
OE
Here Wnc  fd , and power P 
, thus
t
2
(1 / 2)mv 2  fd 0.5 500 kg 110 m/s    1200 N 400 m 
P

t
7.30 s
6
3.505  10 J

 4.80  10 5 W
7.30 s
1 hp
P  4.801  10 5 W 
 643 hp
746 W
38.
(a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp =
746 W) to reach a speed of 15.0 m/s, neglecting friction? (b) How long will this
acceleration take if the car also climbs a 3.00-m-high hill in the process?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 7
W (1 / 2) mv 2

P
P
746 W
P  40.0 hp 
 29,840 W
1 hp
(a) W  Pt  t 
0.5 850 kg 15.0 m/s 
 3.20 s
29,840 W
2
t
2
(b) t  W  (1 / 2)mv  mgh
P
P
2
0.5 850 kg 15.0 m/s   (850 kg) 9.80 m/s 2  3.00 m 

 4.04 s
29,840 W
39.
(a) Find the useful power output of an elevator motor that lifts a 2500-kg load a
height of 35.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note
that the total mass of the counterbalanced system is 10,000 kg—so that only 2500 kg
is raised in height, but the full 10,000 kg is accelerated. (b) What does it cost, if
electricity is $0.0900 per kW h ?
Solution
(a) W  KE  PE 
1 2
mv  mgh
2

W 1  mv 2
P
 
 mgh 
t
t 2

2

1  (10,000 kg) 4.00 m/s 

 (2500 kg) 9.80 m/s 2 (35.0 m) 

12.0 s 
2

4
4
 7.8125  10 W  7.81  10 W
 $0.0900 
(b) cost  78.1 kW 
  $7.03 / h  $0.00195/s
 kW  h 

40.

(a) What is the available energy content, in joules, of a battery that operates a 2.00-W
electric clock for 18 months? (b) How long can a battery that can supply 8.00 10 4 J
run a pocket calculator that consumes energy at the rate of 1.00 10 3 W ?
1y
365 d 24 h 3600 s
Solution (a)
E  Pt  2.00 W 18 mo. 



 9.46 10 7 J
12 mo. 1 y
1d
1h
E 8.00 10 4 J 
(b) E  Pt  t  
 8.00 10 7 s  2.54 y
-3
P 1.00 10 W
OpenStax College Physics
41.
Instructor Solutions Manual
Chapter 7
(a) How long would it take a 1.50  10 5 -kg airplane with engines that produce 100 MW
of power to reach a speed of 250 m/s and an altitude of 12.0 km if air resistance were
negligible? (b) If it actually takes 900 s, what is the power? (c) Given this power, what
is the average force of air resistance if the airplane takes 1200 s? (Hint: You must find
the distance the plane travels in 1200 s assuming constant acceleration.)
Solution (a)
W
W
1
(1 / 2) mv 2  mgh
P
 t  , W  mv 2  mgh  t 
t
P
2
P
2
5
5
(1/2) 1.50  10 kg  250 m/s   1.50  10 kg  (9.80 m/s 2 ) (12.0  10 3 m)
t
1.00  10 8 W
 223 s
223 s
(b) P 
1.00  108 W   2.48  107 W  24.8 MW
900 s
(c) W  1 mv 2  mgh  Ff d . Since , P  W  24.8 MW and t  1200 s,
2
t
6
W  24.8 10 W  1200 s   29.77 10 9 J
v 250 m/s
v  v0  at  a  
 0.208 m/s 2
t
1200 s
1
1
2
d  at 2  0.208 m/s 2  1200 s   150 km
2
2
W  (1 / 2)mv 2  mgh
Ff 
d
2
29.77 10 9 J  1.50 10 5 kg [(. 5)250m/s   (9.80 m/s 2 )(12.0 10 3 m)]

(1.50 10 5 m)
 4.96 10 4 N  49.6 kN
42.
Calculate the power output needed for a 950-kg car to climb a 2.00 slope at a
constant 30.0 m/s while encountering wind resistance and friction totaling 600 N.
Explicitly show how you follow the steps in the Problem-Solving Strategies for Energy.
Solution The energy supplied by the engine is converted into frictional energy as the car goes
up the incline.
W Fd
d 

 F    Fv, where F is parallel to the incline and
t
t
t
F  f  w  600 N  mg sin  . Substituting gives P  ( f  mg sin  )v , so that:
P


P  600 N  (950 kg)(9.80 m/s 2 )sin2  (30.0 m/s)  2.77  10 4 W
OpenStax College Physics
43.
Instructor Solutions Manual
Chapter 7
(a) Calculate the power per square meter reaching Earth’s upper atmosphere from the
Sun. (Take the power output of the Sun to be 4.00 10 26 W.) (b) Part of this is
absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m 2 reaches
Earth’s surface. Calculate the area in km 2 of solar energy collectors needed to replace
an electric power plant that generates 750 MW if the collectors convert an average of
2.00% of the maximum power into electricity. (This small conversion efficiency is due to
the devices themselves, and the fact that the sun is directly overhead only briefly.)
With the same assumptions, what area would be needed to meet the United States’
energy needs (1.05 1020 J)? Australia’s energy needs (5.4 1018 J)? China’s energy
needs (6.3 1019 J)? (These energy consumption values are from 2006.)
Solution (a) The power per meter squared is the power output of the Sun spread over a sphere
P
P
of radius r . Thus
where r is the distance from the Sun to Earth. The

A 4r 2
power of the Sun is given and the average distance from the Sun to Earth is found
in the text. Substituting these values,
P
4.00  10 26 W

 1.4147  10 3 W/m 2  1.41 kW/m 2
11
2
A 4 (1.50  10 m)
(b) Here the power output would be the power per unit area reaching Earth, times
the area, times the efficiency, or P  ( P / A)( A)( Eff ). Solving for the area A ,
P
7.50 108 W

 2.88 10 7 m 2  28.8 km 2 .
( P / A) Eff (1.30 10 3 W/m 2 ) 0.0200
For the United States, the needed power is P  1.05 10 20 J/y  3.32 1012 W .
A
P
3.32 1012 W
A

 1.28 1012 m 2  1.28 10 6 km 2 .
3
2
( P / A) Eff (1.30 10 W/m ) 0.0200
For Australia, the needed power is P  5.4 1018 J/y  1.7 1011W .
P
1.7 1011 W

 6.6 10 9 m 2  6.6 10 3 km 2 .
( P / A) Eff (1.30 10 3 W/m 2 ) 0.0200
For China, the needed power is P  6.3 1019 J/y  2.0 1012 W .
A
P
2.0 1012 W

 7.7 1010 m 2  7.7 10 4 km 2 .
3
2
( P / A) Eff (1.30 10 W/m ) 0.0200
This illustrates the difficulty in replacing a midsize power plant with solar energy.
A
7.8 WORK, ENERGY, AND POWER IN HUMANS
44.
(a) How long can you rapidly climb stairs (116/min) on the 93.0 kcal of energy in a
10.0-g pat of butter? (b) How many flights is this if each flight has 16 stairs?
OpenStax College Physics
Instructor Solutions Manual
Chapter 7
Solution
(a) E  Pt  t 
45.
(a) What is the power output in watts and horsepower of a 70.0-kg sprinter who
accelerates from rest to 10.0 m/s in 3.00 s? (b) Considering the amount of power
generated, do you think a well-trained athlete could do this repetitively for long
periods of time?
Solution
E
93.0 kcal

 9.5 min
P 9.8 kcal/min
116 stairs 1 flight
(b) 9.5 min 

 69 flights
min
16 stairs
E (1 / 2) mv 2 0.5 (70.0 kg) (10.0 m/s) 2


 1.1667  10 3 W  1.17  10 3 W
t
t
3.00 s
1
hp
and P  1.1667  10 3 W 
 1.56 hp
746 W
(b) This is a lot of power generated, so it probably isn’t possible to do this repeatedly
for a long period of time. That’s why long distance runners run slower than
sprinters!
(a) P 


46.
Calculate the power output in watts and horsepower of a shot-putter who takes 1.20 s
to accelerate the 7.27-kg shot from rest to 14.0 m/s, while raising it 0.800 m. (Do not
include the power produced to accelerate his body.)
Solution
net W 
47.
(a) What is the efficiency of an out-of-condition professor who does 2.10  105 J of
useful work while metabolizing 500 kcal of food energy? (b) How many food calories
would a well-conditioned athlete metabolize in doing the same work with an
efficiency of 20%?
Solution
1 2
1
mv  mgh  mv02  mgh0
2
2
1
 (7.27 kg) (14.0 m/s) 2  (7.27 kg) (9.80 m/s 2 ) (0.800 m)  769.5 J
2
W 769.5 J
P

 641.2 W  641 W and
t
1.20 s
1 hp
P  (641.2 W) 
 0.860 hp
746 W
(a) Eff 
Eout
2.10  10 5 J

 0.1004  0.100
Ein
500 kcal 4.184  10 3 J/1kcal
(b) Eff 
Eout
E
1 kcal / 4.184  10 5 J
 Ein  out  (2.10  10 5 J)
 2.51  10 2 kcal
Ein
Eff
0.200


OpenStax College Physics
48.
Solution
Instructor Solutions Manual
Chapter 7
Energy that is not utilized for work or heat transfer is converted to the chemical
energy of body fat containing about 39 kJ/g. How many grams of fat will you gain if
you eat 10,000 kJ (about 2500 kcal) one day and do nothing but sit relaxed for 16.0 h
and sleep for the other 8.00 h? Use data from Table 7.5 for the energy consumption
rates of these activities.
E out  P1t1  P2 t 2  1.7 kcal/min 
60 min
60 min
 16.0 h  1.2 kcal/min 
 8.00 h
1h
1h
 2.208  10 3 kcal
E gained  E in  E out  2500 kcal  2208 kcal  292 kcal
Fat gain  292 kcal 
1g
 31.4 g  31 g
9.3 kcal
49.
Using data from Table 7.5, calculate the daily energy needs of a person who sleeps for
7.00 h, walks for 2.00 h, attends classes for 4.00 h, cycles for 2.00 h, sits relaxed for
3.00 h, and studies for 6.00 h. (Studying consumes energy at the same rate as sitting
in class.)
Solution
Eout  P1t1  P2 t 2  ...
 1.2 kcal/min
420 min   3.8 kcal/min 120 min 
 3.0 kcal/min  (240 min)  (5.7 kcal/min) (120 min)
 (1.7 kcal/min) (180 min)  (3.0 kcal/min) (360 min)
E  3750 kcal  3.8  10 3 kcal
50.
What is the efficiency of a subject on a treadmill who puts out work at the rate of 100
W while consuming oxygen at the rate of 2.00 L/min? (Hint: See Table 7.5.)
Solution
Eff 
51.
Shoveling snow can be extremely taxing because the arms have such a low efficiency
in this activity. Suppose a person shoveling a footpath metabolizes food at the rate of
800 W. (a) What is her useful power output? (b) How long will it take her to lift 3000
kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c)
How much waste heat transfer in kilojoules will she generate in the process?
Solution
(a) Eff 
Eout Pout 100 W


 0.1429  14.3%
Ein
Pin
700 W
Pout
 Pout  Eff  Pin  0.03  800 W  24 W
Pin
W
W mgh 3000 kg  (9.80 m/s 2 ) (12.0 m)
t 


 1.47  10 3 s  24.5 min
t
P
P
24 W
(c) E waste  Pin  Pout t  800 W  24 W  1.47  10 3 s   1.14  10 6 J
(b) P 
OpenStax College Physics
52.
Instructor Solutions Manual
Chapter 7
Very large forces are produced in joints when a person jumps from some height to the
ground. (a) Calculate the force produced if an 80.0-kg person jumps from a 0.600–mhigh ledge and lands stiffly, compressing joint material 1.50 cm as a result. (Be certain
to include the weight of the person.) (b) In practice the knees bend almost
involuntarily to help extend the distance over which you stop. Calculate the force
produced if the stopping distance is 0.300 m. (c) Compare both forces with the weight
of the person.
Solution Given: m  80.0 kg, h  0.600 m, and d  0.0150 m
Find: net F . Using W  Fd and the work-energy theorem gives: W  Fj d  mgh
F
mgh (80.0 kg) (9.80 m/s 2 ) (0.600)

 3.136 104 N.
d
0.0150 m

N

Fj

mg
(a) Now, looking at the body diagram: net F  w  Fj
net F  (80.0 kg) (9.80 m/s 2 )  3.136  10 4 N  3.21 10 4 N
mgh (80.0 kg) (9.80 m/s 2 ) (0.600)

 1568 N.
d
0.300 m
net F  (80.0 kg)(9.80 m/s 2 )  1568 N  2.35  10 3 N
net F 32,144 N
(c) In (a),

 41.0. This could be damaging to the body.
mg
784 N
net F 2352 N
In (b),

 3.00. This can be easily sustained.
mg
784 N
(b) Now, let d  0.300 m so that Fj 
53.
Jogging on hard surfaces with insufficiently padded shoes produces large forces in the
feet and legs. (a) Calculate the force needed to stop the downward motion of a
jogger’s leg, if his leg has a mass of 13.0 kg, a speed of 6.00 m/s, and stops in a
distance of 1.50 cm. (Be certain to include the weight of the 75.0-kg jogger’s body.)
(b) Compare this force with the weight of the jogger.
OpenStax College Physics
Solution
Instructor Solutions Manual
1
mv 2
mv 2  F 
2
2d
2
mv
net F  F  mg 
 mg
2d
2

13.0 kg 6.00 m/s 
net F 
 75.0 kg  9.80 m/s 2
2  0.0150 m
 15,600 N  735 N  1.63  10 4 N
net F 1.634 10 4 N
(b)

 22.2
mg
735 N
(a)
W  Fd 

54.
Chapter 7

(a) Calculate the energy in kJ used by a 55.0-kg woman who does 50 deep knee bends
in which her center of mass is lowered and raised 0.400 m. (She does work in both
directions.) You may assume her efficiency is 20%. (b) What is the average power
consumption rate in watts if she does this in 3.00 min?
Solution (a) Wout  Nmgh  50 (55.0 kg) (9.80 m/s 2 ) (0.800 m)  2.156 10 4 J
Wout
Wout 2.156 10 4 J
Eff 
 Win 

 1.078  105 J  108 kJ
Win
Eff
0.20
5
W
1.078  10 J
(b) P  in 
 599 W
t
180 s
55.
Kanellos Kanellopoulos flew 119 km from Crete to Santorini, Greece, on April 23, 1988,
in the Daedalus 88, an aircraft powered by a bicycle-type drive mechanism (see Figure
7.43). His useful power output for the 234-min trip was about 350 W. Using the
efficiency for cycling from Table 7.2, calculate the food energy in kilojoules he
metabolized during the flight.
Solution
Eff 
Pout
P
350 W
 Pin  out 
 1.75  10 3 W
Pin
Eff
0.20


 60s 
E  Pt  1.75 103 W 234 min  

 1 min 
1 kcal
 2.457 107 J  2.457 107 J
 5.87 103 kcal
3
4.184 10 J

56.

The swimmer shown in Figure 7.44 exerts an average horizontal backward force of
80.0 N with his arm during each 1.80 m long stroke. (a) What is his work output in
each stroke? (b) Calculate the power output of his arms if he does 120 strokes per
minute.
OpenStax College Physics
Instructor Solutions Manual
Chapter 7
Solution (a) W  Fd  (80.0 N)(1.80 m)  144 J
W
120 strokes
144 J


 1.728  10 4 J/min 
(b)
min
1 min
1 strokes
W 1.728  10 4 J
P

 288 W
t
60 s
57.
Mountain climbers carry bottled oxygen when at very high altitudes. (a) Assuming
that a mountain climber uses oxygen at twice the rate for climbing 116 stairs per
minute (because of low air temperature and winds), calculate how many liters of
oxygen a climber would need for 10.0 h of climbing. (These are liters at sea level.)
Note that only 40% of the inhaled oxygen is utilized; the rest is exhaled. (b) How much
useful work does the climber do if he and his equipment have a mass of 90.0 kg and
he gains 1000 m of altitude? (c) What is his efficiency for the 10.0-h climb?
Solution
(a) Oxygen needed 
(b) Wout
1
60min 1.96L
 2  10.0h 

 5.88  103 L
0.40
1h
min
2
 mgh  90.0 kg  9.80 m/s 1000 m   8.82  10 5 J


60 min
60s

 685 W  4.93  10 7 J
1h
1 min
5
8.82  10 J

 0.0179  1.79%
4.93  10 7 J
(c) Win  2  10.0 h 
Eff 
58.
Wout
Win
The awe-inspiring Great Pyramid of Cheops was built more than 4500 years ago. Its
square base, originally 230 m on a side, covered 13.1 acres, and it was 146 m high,
with a mass of about 7  10 9 kg . (The pyramid’s dimensions are slightly different
today due to quarrying and some sagging.) Historians estimate that 20,000 workers
spent 20 years to construct it, working 12-hour days, 330 days per year. (a) Calculate
the gravitational potential energy stored in the pyramid, given its center of mass is at
one-fourth its height. (b) Only a fraction of the workers lifted blocks; most were
involved in support services such as building ramps (see Figure 7.45), bringing food
and water, and hauling blocks to the site. Calculate the efficiency of the workers who
did the lifting, assuming there were 1000 of them and they consumed food energy at
the rate of 300 kcal/h. What does your answer imply about how much of their work
went into block-lifting, versus how much work went into friction and lifting and
lowering their own bodies? (c) Calculate the mass of food that had to be supplied each
day, assuming that the average worker required 3600 kcal per day and that their diet
was 5% protein, 60% carbohydrate, and 35% fat. (These proportions neglect the mass
of bulk and non digestible materials consumed.)
Solution (a) PE  mgh  (7.00  109 kg) (9.80 m/s 2 ) (36.5 m)  2.504  1012 J  2.50  1012 J
OpenStax College Physics
(b) Ein  NPt  1000 
Instructor Solutions Manual
Chapter 7
300 kcal 4186 J
330 d 12 h

 20 y 

 9.946  1013 kcal
h
kcal
y
d
Wout PE 2.504  1012 J


 0.0252  2.52%
Ein
Ein
9.946  1013
(c) Eprotein  (3600 kcal)(0.05 )  180 kcal;
1g
1g
mprotein  Eprotein 
 180 kcal 
 43.90 g;
4.1 kcal
4.1 kcal
Ecarbohydrate  (3600 kcal)(0.60 )  2160 kcal;
Eff 
1g
1g
 2160 kcal 
 526.8 g; and
4.1 kcal
4.1 kcal
 (3600 kcal)(0.35 )  1260 kcal;
mcarbohydrate  Ecarbohydrate 
Efat
1g
1g
 2160 kcal 
 135.5 g;
9.3 kcal
9.3 kcal
Therefore, the total mass of food require for the average worker per day is:
mperson  mprotein  mcarbohydrate  mfat  (43.90 g)  (526.8 g)  (135.5 g)  706.2 g,
mfat  Efat 
and the total amount of food required for the 20,000 workers is:
m  Nmperson  20,000  0.7062 kg  1.4110 4 kg  1.4 10 4 kg
59.
(a) How long can you play tennis on the 800 kJ (about 200 kcal) of energy in a candy
bar? (b) Does this seem like a long time? Discuss why exercise is necessary but may
not be sufficient to cause a person to lose weight.
Solution (a) From Table 7.5: P  6.3 kcal/min for playing tennis. Using E  Pt gives:
E
200 kcal
E  Pt  t  
 31.75 min  32 min
P 6.3 kcal/min
(b) This does seem like a long time, since one candy bar is equivalent to 30 minutes of
tennis. Exercise not only burns calories, but it increases your metabolism. Exercise
is important to losing weight, but watching your calorie intake is also important.
Weight loss only occurs when the amount of energy used is more than the
amount of energy taken in.
7.9 WORLD ENERGY USE
60.
Integrated Concepts (a) Calculate the force the woman in Figure 7.46 exerts to do a
push-up at constant speed, taking all data to be known to three digits. (b) How much
work does she do if her center of mass rises 0.240 m? (c) What is her useful power
output if she does 25 push-ups in 1 min? (Should work done lowering her body be
included? See the discussion of useful work in Work, Energy, and Power in Humans.
OpenStax College Physics
Instructor Solutions Manual
Solution (a) Using  CCW   CW ,
Fr (1.50m)  w (0.900 m)  Fr 
Chapter 7
50.0 kg  9.80 m/s 2 0.900 m   294 N
1.50 m
(b) W  mgh  (50.0 kg)(9.80 m/s )(0.240m)  117.6 J  118 J
W 25 pushups 117.6 J
(c) P 


 49.0 W
t
60 s
pushup
Note: An entirely different process is involved during the time the woman drops
after each push-up. The energy expended during this phase is negligible
compared to the push-up phase.
2
61.
Integrated Concepts A 75.0-kg cross-country skier is climbing a 3.0 slope at a
constant speed of 2.00 m/s and encounters air resistance of 25.0 N. Find his power
output for work done against the gravitational force and air resistance. (b) What
average force does he exert backward on the snow to accomplish this? (c) If he
continues to exert this force and to experience the same air resistance when he
reaches a level area, how long will it take him to reach a velocity of 10.0 m/s?
Solution (a) Beginning with W  Pt , we know:
W Fd
d
P

 F  Fv and FII  f  wII  f  mg sin  .
t
t
t
Therefore, the power is:
P   f  mg sin   v




 25.0 N   75.0 kg  9.80 m/s 2 sin3.00  2.00 m/s   127W
(b) F  f  mg sin   25.0 N  75.0 kg  9.80 m/s 2 sin3.00   63.5 N
(c) If now on a level surface,
F  f 63.5 N  25.0 N

 0.513 m/s 2 .
m
75.0 kg
The initial velocity is v0  2.00 m/s and we want to find the final time. Use
v  v0 10.0 m/s   2.00 m/s 
kinematics: v  v0  at so that t 

 15.6 s.
a
0.513 m/s 2
net FII  ma  F  f so that a 
62.
Integrated Concepts The 70.0-kg swimmer in Figure 7.44 starts a race with an initial
velocity of 1.25 m/s and exerts an average force of 80.0 N backward with his arms
during each 1.80 m long stroke. (a) What is his initial acceleration if water resistance
is 45.0 N? (b) What is the subsequent average resistance force from the water during
the 5.00 s it takes him to reach his top velocity of 2.50 m/s? (c) Discuss whether water
resistance seems to increase linearly with velocity.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 7
80 N   45.0 N   0.500 m/s 2 .
F f

m
75.0 kg
(b) Given t  5.00 s, v0  1.25 m/s, v  2.50 m/s.
Find a , then f . Use kinematics: v  v0  at so that
(a) net Fh  F  f  ma , so that a 
v  v0 2.50 m/s   1.25 m/s 

 0.250 m/s 2
t
5.00 s
Using Newton’s laws: f  F  ma  80.0 N   70.0 kg  0.250 m/s 2   62.5 N
(c) Assuming the acceleration of the swimmer decreases linearly with time over the
5.00 s interval, the frictional force must therefore be increasing linearly with time,
since f  F  ma . If the acceleration decreases linearly with time, the velocity
will contain a term dependent on (t 2 ) . Therefore, the water resistance will not
depend linearly on the velocity.
a
63.
Integrated Concepts A toy gun uses a spring with a force constant of 300 N/m to
propel a 10.0-g steel ball. If the spring is compressed 7.00 cm and friction is negligible:
(a) How much force is needed to compress the spring? (b) To what maximum height
can the ball be shot? (c) At what angles above the horizontal may a child aim to hit a
target 3.00 m away at the same height as the gun? (d) What is the gun’s maximum
range on level ground?
Solution (a) F  kx  300 N/m0.0700 m   21.0 N
kx
300 N/m 0.0700 m   7.50m
1 2
(b) kxi  mghf  hf  i 
2
2mg 20.0100 kg  9.80 m/s 2
2
2

(c)
v0

L =3m
Using the equation R 
v02 sin 2
,
g

OpenStax College Physics

Instructor Solutions Manual
Chapter 7
1  1  Rg  
sin  2  
2 
 v0  
1 2 1 2
kx 2
mv0  kx  v02 
2
2
m
1
 Rg   1  1  Rgm  
  sin 1  2
   sin  2  
2
 kx / m   2 
 kx  

1 1  3.00 m  9.80 m/s 2 0.0100 kg  
sin 

2
300 N/m 0.0700 m 2


  sin 1 0.200    11.54 or
(d) Rmax
64.
1
1
2
2
1
   168.5  5.77 or 84.23
2
2
v0 kx 2 300 N/m 0.0700 m 2



 1.50 m
g
mg 0.0100 kg  9.80 m/s 2


Integrated Concepts (a) What force must be supplied by an elevator cable to produce
an acceleration of 0.800 m/s 2 against a 200-N frictional force, if the mass of the
loaded elevator is 1500 kg? (b) How much work is done by the cable in lifting the
elevator 20.0 m? (c) What is the final speed of the elevator if it starts from rest? (d)
How much work went into thermal energy?
Solution (a) F  mg  Ff  ma 


F  m a  g   Ff  1500 kg  0.800 m/s 2  9.80 m/s 2  200 N  1.6110 4 N
(b) W  Fd  1.6110 N20.0 m  3.22 10 J
(c) v 2  v02  2ax, but v0  0, so that
4
 
5

v  (2ax)1 / 2  2 0.800 m/s 2 (20.0 m)

1/2
 5.66 m/s
(d) Wf  Ff d  200 N  20.0 m  4.00  10 J  4.00 kJ
3
65.
Unreasonable Results A car advertisement claims that its 900-kg car accelerated from
rest to 30.0 m/s and drove 100 km, gaining 3.00 km in altitude, on 1.0 gal of gasoline.
The average force of friction including air resistance was 700 N. Assume all values are
known to three significant figures. (a) Calculate the car’s efficiency. (b) What is
unreasonable about the result? (c) Which premise is unreasonable, or which premises
are inconsistent?
OpenStax College Physics
Solution
Instructor Solutions Manual
1
mv 2  Ff d  mgh
2
2
 0.5 900 kg 30.0 m/s   700 N  100  10 3 m
Chapter 7
(a) Wout 
Wout


 900 kg  9.80 m/s 2 3.00  10 3

m

 9.69  10 J
7
Win  1.3  10 8 J ( from Table 7.1)
Wout 9.69  10 7 J

 74.5%  75%
Win
1.3  10 8 J
(b) Efficiency in excess of 40% for mechanical devices is rare. The representative value
for a gasoline engine from Table 7.2 is 30%.
(c) The mileage rate is most unreasonable. A mileage rate of 100 km/gal is equivalent
to about 62 mi/gal and is generated traveling up a 1.7 grade. This seems unlikely.
Eff 
66.
Unreasonable Results Body fat is metabolized, supplying 9.30 kcal/g, when dietary
intake is less than needed to fuel metabolism. The manufacturers of an exercise
bicycle claim that you can lose 0.500 kg of fat per day by vigorously exercising for 2.00
h per day on their machine. (a) How many kcal are supplied by the metabolization of
0.500 kg of fat? (b) Calculate the kcal/min that you would have to utilize to
metabolize fat at the rate of 0.500 kg in 2.00 h. (c) What is unreasonable about the
results? (d) Which premise is unreasonable, or which premises are inconsistent?
Solution (a) # of kcal  500 g  9.30 kcal/g  4650 kcal  4.65  10 3 kcal
4650 kcal
(b) kcal/min 
 38.75 kcal/min  38.8 kcal/min
120 min
(c) This power output is higher than the highest value in Table 7.5, which is about 35
kcal/min (corresponding to 2415 watts) for sprinting.
(d) It would be impossible to maintain this power output for 2 hours (imagine
sprinting for 2 hours!).
69.
Integrated Concepts A 105-kg basketball player crouches down 0.400 m while waiting
to jump. After exerting a force on the floor through this 0.400 m, his feet leave the
floor and his center of gravity rises 0.950 m above its normal standing erect position.
(a) Using energy considerations, calculate his velocity when he leaves the floor. (b)
What average force did he exert on the floor? (Do not neglect the force to support his
weight as well as that to accelerate him.) (c) What was his power output during the
acceleration phase?
OpenStax College Physics
Instructor Solutions Manual
Solution (a) KEi  PEi  KEf  PEf , so
1
2
mv0  mgh 
2
v0  2 gh 
1/ 2
(b)
 

 2 9.80 m/s 2 0.950 m 

1/2
Chapter 7
 4.315 m/s 2  4.32 m/s
mgh
d
mgh
net F  mg  FJ  mg 
d
105 kg  9.80 m/s 2 0.950 m   3.47  10 3 N
 105 kg  9.80 m/s 2 
0.400m
WJ  FJ d  mgh  FJ 



1/ 2

1
 2h 
3473 N
(c) x  at 2  t    , where a 
 33.07 m/s 2 , so that
2
105 kg
 a 
1/2
 2  0.400 m 
t 
 0.156 s
2 
 33.07 m/s 
W net F  d 3473 N  0.400 m
P 

 8.93 kW
t
t
0.156 s
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