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Potential Energy
CHAPTER OUTLINE
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
Potential Energy of a
System
The Isolated System
Conservative and
Nonconservative Forces
Conservative Forces and
Potential Energy
The Nonisolated System
in Steady State
Potential Energy for
Gravitational and Electric
Forces
Energy Diagrams and
Stability of Equilibrium
Context
ConnectionPotential
Energy in Fuels
ANSWERS TO QUESTIONS
Q7.1
The final speed of the children will not depend on the slide length or
the presence of bumps if there is no friction. If there is friction, a
longer slide will result in a lower final speed. Bumps will have the
same effect as they effectively lengthen the distance over which
friction can do work, to decrease the total mechanical energy of the
children.
Q7.2
Total energy is the sum of kinetic and potential energies. Potential
energy can be negative, so the sum of kinetic plus potential can also
be negative.
Q7.3
Both agree on the change in potential energy, and the kinetic energy. They may disagree on the
value of gravitational potential energy, depending on their choice of a zero point.
Q7.4
(a)
mgh is provided by the muscles.
(b)
No further energy is supplied to the object-Earth system, but some chemical energy must
be supplied to the muscles as they keep the weight aloft.
(c)
The object loses energy mgh, giving it back to the muscles, where most of it becomes
internal energy.
Q7.5
Lift a book from a low shelf to place it on a high shelf. The net change in its kinetic energy is zero,
but the book-Earth system increases in gravitational potential energy. Stretch a rubber band to
encompass the ends of a ruler. It increases in elastic energy. Rub your hands together or let a pearl
drift down at constant speed in a bottle of shampoo. Each system (two hands; pearl and shampoo)
increases in internal energy.
Q7.6
Three potential energy terms will appear in the expression of total mechanical energy, one for each
conservative force. If you write an equation with initial energy on one side and final energy on the
other, the equation contains six potential-energy terms.
Q7.7
(a)
It does if it makes the object’s speed change, but not if it only makes the direction of the
velocity change.
165
166
Potential Energy
(b)
Yes, according to Newton’s second law.
Chapter 7
167
Q7.8
The original kinetic energy of the skidding car can be degraded into kinetic energy of random
molecular motion in the tires and the road: it is internal energy. If the brakes are used properly, the
same energy appears as internal energy in the brake shoes and drums.
Q7.9
All the energy is supplied by foodstuffs that gained their energy from the sun.
Q7.10
Elastic potential energy of plates under stress plus gravitational energy is released when the plates
“slip”. It is carried away by mechanical waves.
Q7.11
The total energy of the ball-Earth system is conserved. Since the system initially has gravitational
energy mgh and no kinetic energy, the ball will again have zero kinetic energy when it returns to its
original position. Air resistance will cause the ball to come back to a point slightly below its initial
position. On the other hand, if anyone gives a forward push to the ball anywhere along its path, the
demonstrator will have to duck.
Q7.12
Kinetic energy is greatest at the starting point. Gravitational energy is a maximum at the top of the
flight of the ball.
Q7.13
Gravitational energy is proportional to mass, so it doubles.
Q7.14
In stirring cake batter and in weightlifting, your body returns to the same conformation after each
stroke. During each stroke chemical energy is irreversibly converted into output work (and internal
energy). This observation proves that muscular forces are nonconservative.
Q7.15
Let the gravitational energy be zero at the lowest point in the
motion. If you start the vibration by pushing down on the
block (2), its kinetic energy becomes extra elastic potential energy
in the spring (U s ). After the block starts moving up at its lower
turning point (3), this energy becomes both kinetic energy (K) and
gravitational potential energy (U g ), and then just gravitational
energy when the block is at its greatest height (1). The energy then
turns back into kinetic and elastic potential energy, and the cycle
repeats.
FIG. Q7.15
Q7.16
Chemical energy in the fuel turns into internal energy as the fuel burns. Most of this leaves the car
by heat through the walls of the engine and by matter transfer in the exhaust gases. Some leaves the
system of fuel by work done to push down the piston. Of this work, a little results in internal
energy in the bearings and gears, but most becomes work done on the air to push it aside. The work
on the air immediately turns into internal energy in the air. If you use the windshield wipers, you
take energy from the crankshaft and turn it into extra internal energy in the glass and wiper blades
and wiper-motor coils. If you turn on the air conditioner, your end effect is to put extra energy out
into the surroundings. You must apply the brakes at the end of your trip. As soon as the sound of
the engine has died away, all you have to show for it is thermal pollution.
Q7.17
A graph of potential energy versus position is a straight horizontal line for a particle in neutral
equilibrium. The graph represents a constant function.
168
Q7.18
Potential Energy
The ball is in neutral equilibrium.
Chapter 7
169
SOLUTIONS TO PROBLEMS
Section 7.1
Potential Energy of a System
P7.1
With our choice for the zero level for potential energy when
the car is at point B,
(a)
UB  0 .
When the car is at point A, the potential energy of the carEarth system is given by
FIG. P7.1
U A  mgy
where y is the vertical height above zero level. With 135 ft  41.1 m , this height is found as:
a f
y  41.1 m sin 40.0 26.4 m .
Thus,
b
ge
ja
f
U A  1 000 kg 9.80 m s 2 26.4 m  2.59  105 J .
The change in potential energy as the car moves from A to B is
U B  U A  0  2.59  10 5 J  2.59  10 5 J .
(b)
With our choice of the zero level when the car is at point A, we have U A  0 . The
potential energy when the car is at point B is given by U B  mgy where y is the vertical
distance of point B below point A. In part (a), we found the magnitude of this distance to be
26.5 m. Because this distance is now below the zero reference level, it is a negative number.
Thus,
b
ge
ja
f
U B  1 000 kg 9.80 m s 2 26.5 m  2.59  105 J .
The change in potential energy when the car moves from A to B is
U B  U A  2.59  10 5 J  0  2.59  10 5 J .
170
P7.2
Potential Energy
(a)
We take the zero configuration of system
potential energy with the child at the
lowest point of the arc. When the string is
held horizontal initially, the initial
position is 2.00 m above the zero level.
Thus,
a fa
f
U g  mgy  400 N 2.00 m  800 J .
(b)
From the sketch, we see that at an angle of
30.0° the child is at a vertical height of
2.00 m 1  cos 30.0 above the lowest
point of the arc. Thus,
a fa
f
a fa
FIG. P7.2
fa
f
U g  mgy  400 N 2.00 m 1  cos 30.0  107 J .
(c)
The zero level has been selected at the lowest point of the arc. Therefore, U g  0 at this
location.
P7.3
The volume flow rate is the volume of water going over the falls each second:
a fb
g
3 m 0.5 m 1.2 m s  1.8 m 3 s
The mass flow rate is
m
V
   1 000 kg m 3 1.8 m 3 s  1 800 kg s
t
t
e
je
j
If the stream has uniform width and depth, the speed of the water below the falls is the same as the
speed above the falls. Then no kinetic energy, but only gravitational energy is available for
conversion into internal and electric energy.
The input power is Pin 
ja f
energy mgy m


gy  1 800 kg s 9.8 m s 2 5 m  8.82  10 4 J s
t
t
t
b
g
b
ge
e
j
The output power is Puseful  efficiency Pin  0.25 8.82  10 4 W  2.20  10 4 W
The efficiency of electric generation at Hoover Dam is about 85% with a head of water (vertical
drop) of 174 m. Intensive research is underway to improve the efficiency of low head generators.
Section 7.2
*P7.4
(a)
The Isolated System
One child in one jump converts chemical energy into mechanical energy in the amount that
her body has as gravitational energy at the top of her jump:
mgy  36 kg 9.81 m s 2 0.25 m  88.3 J. For all of the jumps of the children the energy is
e
ja f
12e
1.05  10 j88.3 J  1.11  10 J .
6
9
Chapter 7
(b)
0.01
1.11  10 9 J  1.11  10 5 J , making the Richter
100
log E  4.8 log 1.11  10 5  4.8 5.05  4.8
magnitude


 0.2 .
1.5
1.5
1.5
The seismic energy is modeled as E 
171
172
P7.5
Potential Energy
(a)
af
U i  Ki  U f  K f :
mgh  0  mg 2 R 
a f af
g 3.50 R  2 g R 
1
mv 2
2
1 2
v
2
v  3.00 gR
(b)
F m
v2
:
R
n  mg  m
v2
R
L
O m L3.00 gR  gO 2.00mg
v
 gP
M
NR P
Q
NR Q M
n  2.00e
5.00  10 kg je
9.80 m s j
n m

2
3
2

FIG. P7.5
 0.098 0 N d own ward
P7.6
(a)

Energy of the particle-Earth system is
conserved as the particle moves between
point P and the apex of the trajectory.
Since the horizontal component of
velocity is constant,

1
1
1
mv i 2  mv ix 2  mv iy 2
2
2
2
1
 mv ix 2  mgh
2
v iy  2(9.80)( 20.0)  19.8 m s
a
fb
FIG. P7.6
ja
ge
f
(b)
K| PB  W g  mg 60.0 m  0.500 kg 9.80 m s 2 60.0 m  294 J
(c)
Now let the final point be point B.
v xi  v xf  30.0 m s
K| PB 
1
1
mv yf 2  mv yi 2  294 J
2
2
af
2
294  v yi 2  1 176  392
m
v yf 2 
v yf  39.6 m s

v B  30.0 m s i  39.6 m s j
b
P7.7
From leaving ground to the highest point,
gb
g
Ki  U i  K f  U f
b
1
m 6.00 m s
2
g 0  0  me9.80 m s jy
2
2
Chapter 7
b6.00 m sg 
y 
2e
9.80 m s j
af
2
The mass makes no difference:
2
1.84 m
173
174
P7.8
Potential Energy
(a)
Energy of the object-Earth system is conserved as the object moves between the release
point and the lowest point. We choose to measure heights from y  0 at the top end of the
string.
eK  U j  eK  U j :
g
g
i
0  mgy i 
f
1
mv 2f  mgy f
2
e9.8 m s ja2 m cos 30f 12 v  e9.8 m s ja2 m f
v  2e
9.8 m s ja
2 m fa
1  cos 30f  2.29 m s
2
2
f
2
2
f
(b)
Choose the initial point at   30 and the final point at   15 :
a
a
f
2 gLa
cos 15 cos 30f  2e
9.8 m s ja
2 m fa
cos 15 cos 30f 
f
0  mg  L cos 30 
vf 
P7.9
1
mv 2f  mg  L cos 15
2
2
1.98 m s
Using conservation of energy for the system of the Earth and the two
objects
(a)
1
g 4.00 m f b
3.00 kg ga
g 4.00 m f a
5.00  3.00fv
b5.00 kg ga
2
2
v  19.6  4.43 m s
(b)
Now we apply conservation of energy for the system of the
3.00 kg object and the Earth during the time interval between
the instant when the string goes slack and the instant at which
the 3.00 kg object reaches its highest position in its free fall.
FIG. P7.9
a f
1
3.00 v 2  mg y  3.00 gy
2
y  1.00 m
y max  4.00 m  y  5.00 m
*P7.10
(a)
aKf
A B
a
  W  W g  mgh  mg 5.00  3.20
f
a fa f
1
1
mv B2  mv A2  m 9.80 1.80
2
2
v B  5.94 m s
a
B
5.00 m
3.20 m
Wg
A C
a f
 mg 3.00 m  147 J
C
2.00 m
f
Similarly, v C  v A2  2 g 5.00  2.00  7.67 m s
(b)
A
FIG. P7.10
Chapter 7
P7.11
(a)

The force needed to hang on is equal to the force F the
trapeze bar exerts on the performer. From the freebody diagram for the performer’s body, as shown,
v2
v2
or F  mg cos   m . Apply
F  mg cos   m


conservation of mechanical energy of the performerEarth system as the performer moves between the
starting point and any later point:

g a
b
f
1
mg    cos  i  mg    cos   mv 2
2
FIG. P7.11
mv 2
Solve for
and substitute into the force equation to

obtain F  mg 3 cos   2 cos  i .
b
(b)
g
At the bottom of the swing,   0 so
b
F  mg 3  2 cos  i
b
g
F  2mg  mg 3  2 cos  i
g
which gives  i  60.0 .
P7.12
The force of tension and subsequent force of compression in
the rod do no work on the ball, since they are perpendicular to
each step of displacement. Consider energy conservation of
the ball-Earth system between the instant just after you strike
the ball and the instant when it reaches the top. The speed at
the top is zero if you hit it just hard enough to get it there.
Ki  U gi  K f  U gf :
af
4a
9.80fa
0.770f
1
mv i2  0  0  mg 2L
2
v i  4 gL 
v i  5.49 m s
L
initial
final
L

vi
FIG. P7.12
175
176
*P7.13
Potential Energy
(a)
The moving sewage possesses kinetic energy in the same amount as it enters and leaves the
pump. The work of the pump increases the gravitational energy of the sewage-Earth
system. We take the equation Ki  U gi  Wpump  K f  U gf , subtract out the K terms, and
choose U gi  0 at the bottom of the sump, to obtain W pump  mgy f . Now we differentiate
through with respect to time:
Ppump 
m
V
gy f  
gy f
t
t
e
 1 050 kg m 3 1.89  10 6 L d
(b)
efficiency 

1 m IF 1 d IF
9.80 m I
G
jF
G
J
G
J
Ka5.49 m f
H1 000 L KH86 400 s KHs J
3
2
1.24  10 3 W
u sefu l ou tp u t work u sefu l ou tp u t work t

total in p u t work
total in p u t work t
mechanical output power 1.24 kW

 0.209  20.9%
input electric power
5.90 kW
The remaining power, 5.90  1.24 kW  4.66 kW is the rate at which internal energy is
injected into the sewage and the surroundings of the pump.
Section 7.3
Conservative and Nonconservative Forces
P7.14
 
W  F  dr and if the force is constant, this can be written as
(a)
z
z d
z ze

  

W  F  dr  F  rf  ri , which depends only on end points, not path.
(b)
 
W  F  dr 
a f
i
j a
je
3 i  4j  dx i  dyj  3.00 N
5.00 m
W  3.00 N x 0
z
5.00 m
f
a
dx  4.00 N
0
a f
5.00 m
 4.00 N y 0
z
5.00 m
f
dy
0
 15.0 J  20.0 J  35.0 J
The same calculation applies for all paths.
P7.15
W OA 
(a)
z
e
z
e
5.00 m
j
dx i  2 y i  x 2 j 
0
and since along this path, y  0
z
5.00 m
2 ydx
0
W OA  0
W AC 
5.00 m
j
dyj  2 y i  x 2 j 
0
For x  5.00 m ,
W AC  125 J
and
W OAC  0  125  125 J
z
5.00 m
2
x dy
0
Chapter 7
continued on next page
177
178
Potential Energy
W OB 
(b)
z
e
z
e
5.00 m
j
dyj  2 y i  x 2 j 
0
since along this path, x  0 ,
x dy
0
W OB  0
W BC 
5.00 m
j
dx i  2 y i  x 2 j 
0
since y  5.00 m ,
z
5.00 m
2
z
5.00 m
2 ydx
0
W BC  50.0 J
W OBC  0  50.0  50.0 J
W OC 
(c)
Since x  y along OC,
W OC 
ze
ze
je
j ze2ydx  x dyj
dx i  dyj  2 y i  x 2 j 
5.00 m
2
j
2 x  x 2 dx  66.7 J
0
(d)
P7.16
F is nonconservative since the work done is path dependent.
Choose the zero point of gravitational potential energy of the object-spring-Earth system as the
configuration in which the object comes to rest. Then because the incline is frictionless, we have
1
EB  E A : K B  U gB  U sB  K A  U gA  U sA or 0  mg d  x sin   0  0  0  kx 2 . Solving for d gives
2
kx 2
d
x .
2mg sin 
a f
P7.17
From conservation of energy for the block-spring-Earth system,
U gt  U si ,
or
1I
b0.250 kg ge9.80 m s jh  F
G
H2 J
Kb5 000 N m ga0.100 m f
2
This gives a maximum height h  10.2 m .
2
FIG. P7.17
Chapter 7
*P7.18
(a)
For a 5-m cord the spring constant is described by F  kx ,
mg  k 1.5 m . For a longer cord of length L the stretch
distance is longer so the spring constant is smaller in
inverse proportion:
a f
5 m mg
 3.33 mg L
L 1.5 m
k
e
K U g Us
j e
i
 K U g Us
j
initial
f
FIG. P7.18
1
0  mgy i  0  0  mgy f  kx 2f
2
mg 2
1 2 1
mg y i  y f  kx f  3.33
xf
2
2
L
d
i
here y i  y f  55 m  L  x f
a
f
1
2
3.33 55.0 m  L
2
55.0 m L  5.04  10 3 m 2  183 m L  1.67L2
55.0 m L 
0  1.67L2  238L  5.04  10 3  0
L
a fe
2a
1.67f
238  238 2  4 1.67 5.04  10 3
j  238  152 
3.33
25.8 m
only the value of L less than 55 m is physical.
(b)
k  3.33
mg
25.8 m
 F  ma
x max  x f  55.0 m  25.8 m  29.2 m
 kx max  mg  ma
3.33
mg
29.2 m  mg  ma
25.8 m
a  2.77 g  27.1 m s 2
P7.19
(a)
U f  Ki  K f  U i
U f  30.0  18.0  10.0  22.0 J
E  40.0 J
(b)
Yes, Emech  K  U is not equal to zero. For conservative forces K  U  0 .
final
179
180
*P7.20
Potential Energy
(a)
g a
b
 b  b 2  4ac 450.8 N 
x

2a

(b)
P7.21
f
Simplified, the equation is 0  9 700 N m x 2  450.8 N x  1 395 N  m . Then
a450.8 N f  4b9 700 N m gb1 395 N  m g
2b
9 700 N m g
2
450.8 N  7 370 N
 0.403 m or  0.357 m
19 400 N m
One possible problem statement: From a perch at a height of 2.80 m above the top of the
pile of mattresses, a 46.0-kg child jumps nearly straight upward with speed 2.40 m/s. The
mattresses behave as a linear spring with force constant 19.4 kN/m. Find the maximum
amount by which they are compressed when the child lands on them. Physical meaning:
The positive value of x represents the maximum spring compression. The negative value
represents the maximum extension of the equivalent spring if the child sticks to the top of
the mattress pile as he rebounds upward without friction.
The distance traveled by the ball from the top of the arc to the bottom is  R . The work done by the
non-conservative force, the force exerted by the pitcher, is E  Fr cos 0 F R . We shall assign
the gravitational energy of the ball-Earth system to be zero with the ball at the bottom of the arc.
1
1
1
1
Then Emech  mv 2f  mv i2  mgy f  mgy i becomes mv 2f  mv i2  mgy i  F R or
2
2
2
2
af
af
v f  v i2  2 gy i 
a f a15.0f  2a9.80fa1.20f 2a30.0f a0.600f
0.250
2 F R

m
2
v f  26.5 m s
*P7.22
(a)
Maximum speed occurs
after the needle leaves the
spring, before it enters the
body.
v max
FIG. P7.22(a)
Ki  U i  f k d  K f  U f .
We assume the needle is fired horizontally
1 2
1
2
kx  0  mv max
0
2
2
1
1
2
2
375 N m 0.081 m  0.005 6 kg v max
2
2
0
b
ga f b
g
F2a1.23 Jf I  v  21.0 m s
G
H0.005 6 kg J
K
12
max
Chapter 7
continued on next page
181
182
Potential Energy
(b)
The same energy of 1.23 J as in part (a) now becomes partly internal energy in the soft
tissue, partly internal energy in the organ, and partly kinetic energy of the needle just
before it runs into the stop. We make up a work-energy equation to describe this process:
vf
FIG. P7.22(b)
K i  U i  f k 1 d1  f k 2 d 2  K f  U f
1
1 2
1
kx  f k1 d1  f k 2 d 2  mv 2f  0
2
2
a
a
f
f 12 b0.005 6 kg gv
1.23 J  7.60 N 0.024 m  9.20 N 0.035 m 
F2a1.23 J  0.182 J  0.322 JfI  v
G
J
H 0.005 6 kg
K
2
f
12
P7.23
U i  Ki  Emech  U f  K f :
m 2 gh  fh 
f
 16.1 m s
1
1
m1v 2  m 2 v 2
2
2
f  n  m 1 g
b g
2b
m  m g
bhgg

m 2 gh  m 1 gh 
v2
v
P7.24
2
1
m1  m 2 v 2
2
1
FIG. P7.23
m1  m 2
e
ja
f
b
2 9.80 m s 2 1.50 m 5.00 kg  0.400 3.00 kg
8.00 kg
v f  6.20 m s
(1)
We recognize that the change in potential energy is of
two types: gravitational and chemical potential energy
stored in the body of the child from past meals. We also
recognize that the change in internal energy will be due
to the friction force (air resistance and rolling resistance)
as the child rolls down the hill. Therefore, we write (1)
as,
K  U g  U body  fk d  0
3.74 m s
v i  1.40 m s
We begin with Equation 6.20 for the isolated childwheelchair-Earth system,
K  U  Eint  0
g
(2)
12.4 m
FIG. P7.24
2.60 m
Chapter 7
continued on next page
183
184
Potential Energy
The change in potential energy in the body of the child will be due to work done within the system
by the child on the wheels of the chair. Consequently, the requested answer, the work done by the
child, is W child  U body . Therefore, (2) can be expressed as,
1
1
F
I  dmgh  mgh i f d
G
H2 mv  2 mv J
K
1
 me
v  v j  mgd
h  h i f d
2
1
 b
47.0 kg g
b6.20 m sg b1.40 m sg  b47.0 kg ge9.80 m s ja0  2.60 m f a41.0 N fa12.4 m f
2
W ch ild  K  U g  f k d 
2
f
2
f
2
i
f
i
2
i
f
i
k
k
2
2
2
 168 J
P7.25
e
j
a
f
K 
(b)
U  mg 3.00 m sin 30.0  73.5 J
(c)
The mechanical energy converted due to friction is 86.5 J
f
(d)
FIG. P7.25
86.5 J
 28.8 N
3.00 m
f   k n   k mg cos 30.0  28.8 N
k 
P7.26
1
1
m v 2f  v i2   mv i2  160 J
2
2
(a)
28.8 N
b5.00 kg ge9.80 m s jcos 30.0
2
 0.679
Consider the whole motion: Ki  U i  Emech  K f  U f
(a)
0  mgy i  f1 x 1  f 2 x 2 
1
mv 2f  0
2
b80.0 kg ge9.80 m s j1 000 m  a50.0 N fa800 m f b3 600 N ga200 m f 12 b80.0 kg gv
1
784 000 J  40 000 J  720 000 J  b
80.0 kg g
v
2
2b
24 000 Jg
v 
 24.5 m s
2
2
f
2
f
f
(b)
(c)
80.0 kg
Yes this is too fast for safety.
Now in the same energy equation as in part (a), x 2 is unknown, and x 1  1 000 m  x 2 :
Chapter 7
a
fb
gb g
784 000 J  50 000 J  b
3 550 N g
x  1 000 J
784 000 J  50.0 N 1 000 m  x 2  3 600 N x 2 
2
x 2 
continued on next page
733 000 J
 206 m
3 550 N
b
gb
1
80.0 kg 5.00 m s
2
g
2
185
186
Potential Energy
(d)
P7.27
(a)
Really the air drag will depend on the skydiver’s speed. It will be larger than her 784 N
weight only after the chute is opened. It will be nearly equal to 784 N before she opens the
chute and again before she touches down, whenever she moves near terminal speed.
aK  Uf  E
mech
i
0
a f
 K U f :
1 2
1
kx  fx  mv 2  0
2
2
b
ge
1
8.00 N m 5.00  10 2 m
2
v
(b)
e
j
2
j  e3.20  10
2
ja
f 12 e5.30  10
N 0.150 m 
3
j
kg v 2
2 5.20  10 3 J
5.30  10
3
kg
1.40 m s
When the spring force just equals the friction force, the ball will stop speeding up. Here

Fs  kx ; the spring is compressed by
3.20  10 2 N
 0.400 cm
8.00 N m
and the ball has moved
5.00 cm  0.400 cm  4.60 cm from the start.
(c)
Between start and maximum speed points,
1 2
1
1
kx i  fx  mv 2  kx 2f
2
2
2
2
1
1
1
8.00 5.00  10 2  3.20  10 2 4.60  10 2  5.30  10 3 v 2  8.00 4.00  10 3
2
2
2
v  1.79 m s
e
j e
je
j e
j
e
j
2
Chapter 7
P7.28
 Fy  n  mg cos 37.0  0
 n  mg cos 37.0  391 N
a f
f  n  0.250 391 N  97.8 N
 fx  Emech
a97.8fa20.0f U  U  K  K
U  m gd
h  h i a
50.0fa
9.80fa
20.0 sin 37.0f 5.90  10
U  m gd
h  h i a
100fa
9.80fa
20.0f 1.96  10
1
K  m e
v v j
2
m
1
K  m e
v  v j
K  2 K
2
m
A
A
A
B
B
f
B
A
B
4
f
A
A
B
B
3
i
i
2
f
2
f


2
i
2
i
B
A
A
A


Adding and solving, K A  3.92 kJ .
FIG. P7.28
187
188
P7.29
Potential Energy
(a)
The object moved down distance 1.20 m  x . Choose y  0 at its lower point.
K i  U gi  U si  Emech  K f  U gf  U sf
1 2
kx
2
0  mgy i  0  0  0  0 
b1.50 kg ge9.80 m s ja1.20 m  x f 12 b320 N m gx
0b
160 N m g
x a
14.7 N fx  17.6 J
14.7 N  a
14.7 N f  4b
160 N m g
a17.6 N  m f
x
2b
160 N m g
2
2
2
2
x
14.7 N  107 N
320 N m
The negative root tells how high the object will rebound if it is instantly glued to the spring.
We want
x  0.381 m
(b)
From the same equation,
b1.50 kg ge1.63 m s ja1.20 m  x f 12 b320 N m gx
2
2
0  160 x 2  2.44x  2.93
The positive root is x  0.143 m .
(c)
The equation expressing the energy version of the nonisolated system model has one more
term:
mgy i  fx 
1 2
kx
2
b1.50 kg ge9.80 m s ja1.20 m  x f 0.700 N a1.20 m  x f 12 b320 N m gx
2
17.6 J  14.7 N x  0.840 J  0.700 N x  160 N m x 2
160 x 2  14.0 x  16.8  0
x
14.0 
a14.0f  4a160fa16.8f
x  0.371 m
2
320
2
Chapter 7
P7.30
The total mechanical energy of the skysurfer-Earth system is
1
mv 2  mgh.
2
Emech  K  U g 
Since the skysurfer has constant speed,
af
dEmech
dv
dh
 mv
 mg
 0  mg  v  mgv .
dt
dt
dt
The rate the system is losing mechanical energy is then
b
ge
jb
g
dEmech
 mgv  75.0 kg 9.80 m s 2 60.0 m s  44.1 kW .
dt
Section 7.4
P7.31
(a)
Conservative Forces and Potential Energy
z
W  Fx dx 
2x
za2x  4fdx  FG
H2
5.00 m
1
(b)
K  U  0
(c)
K  K f 
(a)
U    Ax  Bx 2 dx 
ze
K f  K 
j
0
(b)
U  
z
3.00 m
Fdx 
P7.33
P7.34
af
5.00 m
 25.0  20.0  1.00  4.00  40.0 J
1
mv 12
 62.5 J
2
Ax 2 Bx 3

2
3
e ja f
A 3.00 2  2.00
2.00 m
K 
I
 4x J
K
U   K  W  40.0 J
mv 12
2
x
P7.32
2
2
2

a f  a2.00f
B 3.00
3
3
3

5.00
19.0
A
B
2
3
5.00
19.0 I
F
G
H 2 A  3 BJ
K
U r 
A
r
Fr  
U
d A
A

 2 . The positive value indicates a force of repulsion if A is positive.
r
dr r
r
F
I
G
HJ
K
e
j e
e
3x y  7x j
U
F 

 e
3x
 3x 3 y  7x
U
Fx  

  9x 2 y  7  7  9x 2 y
x
x
j
3
y
y
y
3
j
 0  3 x 3

Thus, the force acting at the point x , y is F  Fx i  Fy j 
b g
e7  9x y ji  3x j .
2
3
189
190
Potential Energy
Chapter 7
Section 7.5
191
The Nonisolated System in Steady State
No problems in this section
Section 7.6
Potential Energy for Gravitational and Electric Forces
Note: Problems 20.10, 20.13, 20.18, 20.19, 20.20, and 20.62 in Chapter 20 can be assigned with Section 7.6.
P7.35
(a)
U 
e
jb
je
f
g
6.67  10 11 N  m 2 kg 2 5.98  10 24 kg 100 kg
GM E m

 4.77  10 9 J
r
6.37  2.00  10 6 m
a
(b), (c) Planet and satellite exert forces of equal magnitude on each other, directed downward on
the satellite and upward on the planet.
F
P7.36
U  G
GM E m
r
2
e6.67  10

jb
je
11
N  m 2 kg 2 5.98  10 24 kg 100 kg
2
e8.37  10 m j
6
569 N
GM E
Mm
and g 
so that
r
RE 2
F
1
1 I 2

G
H3R R J
K 3 mgR
2
U  b
1000 kg g
e9.80 m s je6.37  10 m j 
3
U  GMm
E
E
E
2
P7.37
g
6
4.17  1010 J
The height attained is not small compared to the radius of the Earth, so U  mgy does not apply;
GM 1 M 2
does. From launch to apogee at height h,
U 
r
Ki  U i  Emech  K f  U f :
GM E M p
GM E M p
1
M p vi 2 
00
2
RE
RE  h
1
10.0  10 3 m s
2
e
j  e6.67  10
2
e
11
  6.67  10 11 N  m 2 kg 2
e5.00  10
7
F5.98  10
jG
H6.37  10
F5.98  10 kg I
jG
H6.37  10 m  hJ
K
je
h  2.52  107 m
3.99  10 14 m 3 s 2
1.26  10 7 m 2 s 2
6
I
m J
K
kg
24
6
j
m 2 s 2  6.26  10 7 m 2 s 2 
6.37  10 6 m  h 
24
N  m 2 kg 2
3.99  10 14 m 3 s 2
6.37  10 6 m  h
 3.16  10 7 m
192
Potential Energy
Chapter 7
P7.38
(a)
F
Gm m I
G
H r J
K
kg je
5.00  10 kg j
U Tot  U 12  U 13  U 23  3U 12  3 
1
193
2
12
U Tot  
(b)
e
3 6.67  10 11 N  m 2
3
2
2
0.300 m
 1.67  10 14 J
At the center of the equilateral triangle
Section 7.7
Energy Diagrams and Stability of Equilibrium
P7.39
(a)
Fx is zero at points A, C and E; Fx is positive at point B and negative at point D.
(b)
A and E are unstable, and C is stable.
(c)
Fx
B
A
C
E
x (m)
D
FIG. P7.39(c)
P7.40
(a)
There is an equilibrium point wherever the graph of potential energy is horizontal:
At r  1.5 mm and 3.2 mm, the equilibrium is stable.
At r  2.3 mm , the equilibrium is unstable.
A particle moving out toward r   approaches neutral equilibrium.
(b)
The system energy E cannot be less than –5.6 J. The particle is bound if 5.6 J  E  1 J .
(c)
If the system energy is –3 J, its potential energy must be less than or equal to –3 J. Thus, the
particle’s position is limited to 0.6 mm  r  3.6 mm .
(d)
K  U  E . Thus, K max  E  U min  3.0 J  5.6 J  2.6 J .
(e)
Kinetic energy is a maximum when the potential energy is a minimum, at r  1.5 mm .
(f)
3 J  W  1 J. Hence, the binding energy is W  4 J .
a f
194
P7.41
Potential Energy
(a)
When the mass moves distance x, the length of each spring
x 2  L2 , so each exerts force
changes from L to
k
F
Hx
2
IK
 L2  L towards its fixed end. The y-components
cancel out and the x components add to:
Fx  2 k
F
Hx
I  2kx 
x
IKF
G
Hx  L J
K
 L2  L
2
2
2 kLx
x 2  L2
2
FIG. P7.41(a)
Choose U  0 at x  0 . Then at any point the potential energy of the system is
2 kLx I
x
2 kx 
dx  2 k z
xdx  2 kL z
J
z zFG
H x L K
x L
U af
x  kx  2 kLF
HL  x  L IK
af
x
x
0
0
U x   Fx dx  
2
(b)
2
2
2
x
x
0
0
2
2
dx
2
F
IK
H
For negative x, U af
x has the same value as for
af
U x  40.0 x 2  96.0 1.20  x 2  1.44
positive x. The only equilibrium point (i.e., where
Fx  0 ) is x  0 .
(c)
Ki  U i  Emech  K f  U f
b
g
1
1.18 kg v 2f  0
2
v f  0.823 m s
0  0.400 J  0 
Section 7.8
*P7.42
FIG. P7.41(b)
Context ConnectionPotential Energy in Fuels
v  100 km h  27.8 m s
The retarding force due to air resistance is
R
a fe
je
Comparing the energy of the car at two points along the hill,
Ki  U gi  E  K f  U gf
or
jb
1
1
DAv 2  0.330 1.20 kg m 3 2.50 m 2 27.8 m s
2
2
af
K i  U gi  W e  R s  K f  U gf
where W e is the work input from the engine. Thus,
g 382 N
2
Chapter 7
af d
ie
W e  R s  K f  K i  U gf  U gi
continued on next page
j
195
196
Potential Energy
Recognizing that K f  Ki and dividing by the travel time t gives the required power input from
the engine as
y I
W I
F
Fs IJ mgF
G
Ht K G
Ht J
K RG
Ht J
K Rv  mgv sin 
Pa
382 N fb
27.8 m s gb
 1 500 kg g
e9.80 m s jb27.8 m sgsin 3.20
P
e
2
P  33.4 kW  44.8 hp
*P7.43
The energy per mass values are:
gasoline:
44 MJ kg
hydrogen:
142 MJ kg , larger by
hay:
17 MJ kg , smaller than gasoline by
battery:
energy Pt 1 200 W 3 600 s
17


 0.27 MJ kg , smaller than hay by
 63 times.
mass
m
16 kg
0.27
142
 3.2 times
44
b
44
 2.6 times
17
g
Thus in the order requested, the energy densities are
0.27 MJ kg for a battery
17 MJ kg for h ay is 63 times larger
44 MJ kg for gasolin e is 2.6 times larger still
142 MJ kg for h yd rogen is 3.2 times larger still
*P7.44
(a)
(b)
b gF
I
H J
K
a fa fa f e j G
Fgal IJ. Its rate of energy conversion is
The car uses gasoline at the rate 55 m i h G
H25 m i K
F2.54 kg Ib55 mi h gF
gal IF 1 h I
P  44  10 J kg G
G
J
H25 mi J
KG
H1 gal K
H3 600 s J
K 6.83  10 W . Its power-per600  10 3 J s h 1 d
energy
P
600 kWh



 6.75 W m 2
area t  area
30 d 13 m 9.5 m
30 d 123.5 m 2 24 h
6
footprint-area is
(c)
4
P
6.83  10 4 W

 6.64  10 3 W m 2 .
area 2.10 m 4.90 m
a
f
For an automobile of typical weight and power to run on sunlight, it would have to carry a
solar panel huge compared to its own size. Rather than running a conventional car, it is
much more natural to use solar energy for agriculture, lighting, space heating, water
heating, and small appliances.
197
Chapter 7
Additional Problems
P7.45
At a pace I could keep up for a half-hour exercise period, I climb two stories up, traversing forty
steps each 18 cm high, in 20 s. My output work becomes the final gravitational energy of the system
of the Earth and me,
b ge
ja
f
mgy  85 kg 9.80 m s 2 40  0.18 m  6 000 J
making my sustainable power
P7.46
6 000 J
 ~ 10 2 W .
20 s
m  mass of pumpkin
R  radius of silo top


v2
 Fr  mar  n  mg cos   m R




When the pumpkin first loses contact with the surface, n  0 .
Thus, at the point where it leaves the surface: v 2  Rg cos  .
FIG. P7.46
Choose U g  0 in the   90.0 plane. Then applying conservation of energy for the pumpkin-Earth
system between the starting point and the point where the pumpkin leaves the surface gives
K f  U gf  Ki  U gi
1
mv 2  mgR cos   0  mgR
2
Using the result from the force analysis, this becomes
1
mRg cos   mgR cos   mgR , which reduces to
2
cos  
bg
2
, and gives   cos 1 2 3  48.2
3
as the angle at which the pumpkin will lose contact with the surface.
*P7.47
When block B moves up by 1 cm, block A moves down by 2 cm and the separation becomes 3 cm.
v
h
We then choose the final point to be when B has moved up by and has speed A . Then A has
3
2
2h
moved down
and has speed v A :
3
198
Potential Energy
eK
A
 KB  U g
000
j  eK
A
i
 KB  U g
j
f
F
IJ  mgh  mg 2h
G
HK 3 3
2
v
1
1
mv A2  m A
2
2
2
mgh 5
 mv A2
3
8
8 gh
15
vA 
P7.48
b
ja
ge
f
(a)
U A  mgR  0.200 kg 9.80 m s 2 0.300 m  0.588 J
(b)
K A  U A  KB  U B
K B  K A  U A  U B  mgR  0.588 J
vB 
(d)
U C  mghC  0.200 kg 9.80 m s 2 0.200 m  0.392 J
FIG. P7.48
ge
ja f
 K  U  U  mgb
h h g
b
0.200 kg g
e9.80 m s ja0.300  0.200fm 
b
KC
KC
P7.49
a f
2 0.588 J
2K B

 2.42 m s
m
0.200 kg
(c)
A
A
C
A
C
2
b
gb
1
1
mv B2  0.200 kg 1.50 m s
2
2
g
2
(a)
KB 
(b)
Em ech  K  U  K B  K A  U B  U A
0.196 J
0.225 J
b g
 0.225 J  b
0.200 kg g
e9.80 m s ja0  0.300 m f
 K B  mg hB  hA
2
 0.225 J  0.588 J  0.363 J
(c)
P7.50
It’s possible to estimate an effective coefficient of friction, but not to find the actual value of
 since n and f vary with position.
k  2.50  10 4 N m ,
m  25.0 kg
x A  0.100 m ,
Ug
(a)
Emech  K A  U gA  U sA
x 0
 Us
x 0
0
Emech  0  mgx A 
1 2
kx A
2
Chapter 7
b ge
1
 e
2.50  10
2
199
f
ja
N m ja
0.100 m f
Emech  25.0 kg 9.80 m s 2 0.100 m
4
2
Emech  24.5 J  125 J  100 J
(b)
Since only conservative forces are involved, the total energy of the child-pogo-stick-Earth
system at point C is the same as that at point A.
KC  U gC  U sC  K A  U gA  U sA :
b
ge
x C  0.410 m
continued on next page
j
0  25.0 kg 9.80 m s 2 x C  0  0  24.5 J  125 J
200
Potential Energy
(c)
b
a
g
f
1
25.0 kg v B2  0  0  0  24.5 J  125 J
2
K B  U gB  U sB  K A  U gA  U sA :
v B  2.84 m s
(d)
K and v are at a maximum when a   F m  0 (i.e., when the magnitude of the upward
spring force equals the magnitude of the downward gravitational force).
This occurs at x  0 where
k x  mg
or
x 
Thus,
K  K max at x  9.80 mm
b25.0 kg ge9.8 m s j  9.80  10
2
(e)
e
Kmax  K A  U gA  U g
b
j eU
sA
g
 Us
b
x 9.80 mm
ge

1
2.50  10 4 N m
2
e
m
j
1
2
25.0 kg v max
 25.0 kg 9.80 m s 2
2
or
j a0.100 mf b0.009 8 mg
j a0.100 mf  b0.009 8 mg
2
2
v max  2.85 m s
yielding
P7.51
x 9.80 mm
3
2.50  10 4 N m
Em ech   fd
E f  Ei   f  d BC
1 2
kx  mgh   mgd BC
2
mgh  12 kx 2

 0.328
mgd BC
P7.52
FIG. P7.51
(a)

d
F 
 x 3  2 x 2  3 x i 
dx
(b)
F 0
e
j e3x
2
j
 4x  3 i
when x  1.87 and  0.535
(c)
The stable point is at
af
x  0.535 point of minimum U x .
FIG. P7.52
The unstable point is at
af
x  1.87 maximum in U x .
Chapter 7
P7.53
aK  U f  aK  U f
0b
30.0 kg g
e9.80 m s ja0.200 m f 12 b250 N m ga0.200 m f
1
 b
50.0 kg g
v b
20.0 kg g
e9.80 m s ja0.200 m fsin 40.0
2
58.8 J  5.00 J  b
25.0 kg g
v  25.2 J
i
f
2
2
2
2
2
FIG. P7.53
v  1.24 m s
P7.54
(a)
Between the second and the third picture, Emech  K  U
1
1
 mgd   mv i2  kd 2
2
2
b
g
b
ge
b
gb
1
1
50.0 N m d 2  0.250 1.00 kg 9.80 m s 2 d  1.00 kg 3.00 m s
2
2
2.45  21.35 N
d
 0.378 m
50.0 N m
(b)
j
g 0
2
Between picture two and picture four, Emech  K  U
a f 12 mv  12 mv
2
v b
3.00 m s g
2 0.378 m f
b1.00 kg ga2.45 N fafa
 f 2d 
2
2
i

2
 2.30 m s

(c)
For the motion from picture two to picture five,
Emech  K  U
a
f
b
gb
g
a
1
2
1.00 kg 3.00 m s
2
9.00 J
D
 2 0.378 m  1.08 m
2 0.250 1.00 kg 9.80 m s 2
 f D  2d  
a fb
P7.55
(a)
ge
f
FIG. P7.54
1 2 1
kx  mv 2
2
2

ga f  12 b0.500 kg gb12.0 m sg

Initial compression of spring:
b
j

1
450 N m x
2
 x  0.400 m
2
2
FIG. P7.55
201
202
Potential Energy
continued on next page
203
Chapter 7
(b)
Speed of block at top of track: Emech   fd
1
1
F
I F
I   f aR f
G
Hmgh  2 mv J
KG
Hmgh  2 mv J
K
b0.500 kg ge9.80 m s ja2.00 m f 12 b0.500 kg gv  12 b0.500 kg gb12.0 m sg
 a
7.00 N fafa
 1.00 m f
T
2
T
2
B
B
2
2
2
T
0.250 v T2  4.21
 v T  4.10 m s
(c)
Does block fall off at or before top of track? Block falls if ac  g
a f
2
4.10
v2
ac  T 
 16.8 m s 2
R
1.00
Therefore ac  g and the block stays on the track .
P7.56
Let  represent the mass of each one meter of the chain and T
represent the tension in the chain at the table edge. We imagine the
edge to act like a frictionless and massless pulley.








(a)
For the five meters on the table with motion impending,
 Fy  0 :
n  5g  0
n  5g
bg

fs   sn  0.6 5g  3g
 Fx  0 :
T  f s  0
T  fs
T  3g
FIG. P7.56
The maximum value is barely enough to support the hanging segment according to
 Fy  0 :
T  3g  0
T  3g
so it is at this point that the chain starts to slide.
(b)
Let x represent the variable distance the chain has slipped since the start.
a f
n  a
5  x fg  0
n a
5  x fg
f   n  0.4a
5  x fg  2g  0.4xg
Then length 5  x remains on the table, with now
 Fy  0 :
k
k
Consider energies of the chain-Earth system at the initial moment when the chain starts to
slip, and a final moment when x  5 , when the last link goes over the brink. Measure
heights above the final position of the leading end of the chain. At the moment the final
link slips off, the center of the chain is at y f  4 meters.
Originally, 5 meters of chain is at height 8 m and the middle of the dangling segment is at
3
height 8   6.5 m .
2
204
Potential Energy
continued on next page
Chapter 7
1
gz F
G
H2 mv
f
b
Ki  U i  Emech  K f  U f :
0  m 1 gy 1  m 2 gy 2 i  f k dx 
2
 mgy
205
IJ
K
f
i
b5gg8  b3gg6.5  zb2g  0.4xggdx  12 b8 gv  b8gg4
5
2
0
z
z
5
5
40.0 g  19.5 g  2.00 g dx  0.400 g x dx  4.00 v 2  32.0 g
0
5
27.5 g  2.00 gx 0  0.400 g
a f
0
2 5
x
2
 4.00 v 2
0
a f
27.5 g  2.00 g 5.00  0.400 g 12.5  4.00 v 2
22.5 g  4.00 v
2
a22.5 m fe9.80 m s j 
2
v
P7.57
4.00
7.42 m s
b
g
The geometry reveals D  L sin   L sin  , 50.0 m  40.0 m sin 50 sin  ,   28.9
(a)
From takeoff to alighting for the Jane-Earth system
eK  U j  W  eK  U j
1
mv  mga
 L cos  f FD a
1f 0  mgb
 L cos  g
2
1
50 kg v  50 kg e
9.8 m s ja
40 m cos 50f 110 N a
50 m f 50 kg e
9.8 m s ja
40 m cos 28.9f
2
g
win d
i
g
f
2
i
2
i
2
2
1
50 kg v i2  1.26  10 4 J  5.5  10 3 J  1.72  10 4 J
2
vi 
(b)
a f
2 947 J
 6.15 m s
50 kg
For the swing back
a f
g af
f
e
ja
 130 kg e
9.8 m s ja
40 m cos 50f
b
1
mv i2  mg  L cos   FD 1  0  mg  L cos 
2
1
130 kg v i2  130 kg 9.8 m s 2 40 m cos 28.9  110 N 50 m
2
2
1
130 kg v i2  4.46  10 4 J  5 500 J  3.28  10 4 J
2
vi 
b
g
2 6 340 J
130 kg
9.87 m s
a f
206
P7.58
Potential Energy
1
1
mv 2  kx 2
2
2
Case I: Surface is frictionless
b
gb
5.00 kg 1.20 m s
mv 2
k 2 
x
10 2 m 2
g  7.20  10
2
2
N m
 k  0.300
Case II: Surface is rough,
1
1
mv 2  kx 2   k mgx
2
2
5.00 kg 2 1
v  7.20  10 2 N m 10 1 m
2
2
v  0.923 m s
e
*P7.59
(a)
je
j  a0.300fb5.00 kg ge9.80 m s je10
2
2
1
m
j
eK  U j  eK  U j
g
g
A
0  mgy A 
B
1
mv B2  0
2
b
(b)
11.1 m s
v2
ac 

r
6.3 m
(c)
 Fy  may
e
j
v B  2 gy A  2 9.8 m s 2 6.3 m  11.1 m s
g
2
19.6 m s 2 up
n B  mg  mac
e
j
n B  76 kg 9.8 m s 2  19.6 m s 2  2.23  103 N up
(d)
a
f
W  Fr cos   2.23  103 N 0.450 m cos 0  1.01  103 J
This quantity of work represents chemical energy, previously in the skateboarder’s body,
that is converted into extra kinetic energy:
eK  U
g
 U chemical
j  eK  U
B
g
 U chemical
j
C
b
g
ja
1
mv B2  0  U chemical, B  U chemical, C  K C  mg y C  y B
2
1
2
76 kg 11.1 m s  1.01  10 3 J  76 kg 9.8 m s 2 0.450 m  K C
2
b gb
(e)
g
e
eK  U j  eK  U j
g
g
C
5.36  10 3 J 
e5.36  10
3
D
b g
ja
1
76 kg v D2  76 kg 9.8 m s 2 5.85 m
2
e
j
J  4.36  10 3 J 2
76 kg
 v D  5.14 m s
f
f
Chapter 7
continued on next page
207
208
Potential Energy
(f)
eK  U j  eK  U j
g
D
g
E
where E is the apex of his motion
b
1
mv D2  0  0  mg y E  y D
2
(g)
b
e
g
j
2
5.14 m s
v2
yE  yD  D 
 1.35 m
2 g 2 9.8 m s 2
g
Consider the motion with constant acceleration between takeoff and touchdown. The time
is the positive root of
y f  y i  v yi t 
1
ay t 2
2
2.34 m  0  5.14 m s t 
1
9.8 m s 2 t 2
2
e
j
4.9t 2  5.14t  2.34  0
t
P7.60
a fa f 
5.14  5.14 2  4 4.9 2.34
9.8
1.39 s
If the spring is just barely able to lift the lower block from the table, the spring lifts it through no
noticeable distance, but exerts on the block a force equal to its weight Mg. The extension of the

spring, from Fs  kx , must be M g k . Between an initial point at release and a final point when the
moving block first comes to rest, we have
4mg I 1 F
4mg I
Mg I
F
FMg IJ 1 kF
G
Hk J
K 2 kG
Hk J
K 0  mgG
Hk K 2 G
Hk J
K
2
Ki  U gi  U si  K f  U gf  U sf :
2
0  mg 
4m 2 g 2 8m 2 g 2 mMg 2 M 2 g 2



k
k
k
2k
2
M
4m 2  mM 
2
M2
 mM  4m 2  0
2

che4m j  m 
M
2ch
3  1f 2m .
Only a positive mass is physical, so we take M  ma
m  m 2  4
1
2
1
2
2
9m 2
Chapter 7
*P7.61
(a)
(b)
Energy is conserved in the swing of the pendulum,
and the stationary peg does no work. So the ball’s
speed does not change when the string hits or leaves
the peg, and the ball swings equally high on both
sides.

L
d
Peg
Relative to the point of suspension, U i  0 ,
a f
U f  mg d  L  d . From this we find that
1
mg 2d  L  mv 2  0 . Also for centripetal motion,
2
mv 2
where R  L  d . Upon solving, we get
mg 
R
3L
.
d
5
a f
P7.62
(a)
FIG. P7.61
At the top of the loop the car and riders are in
free fall:

mv 2
mg down 
down
R
 Fy  may :

v  Rg
Energy of the car-riders-Earth system is
conserved between release and top of loop:
Ki  U gi  K f  U gf :
0  mgh 

af
1
mv 2  mg 2 R
2
af
1
Rg  g 2 R
2
h  2.50 R
gh 
(b)
Let h now represent the height  2.5R of the
release point. At the bottom of the loop we have
mgh 
1
mv b2
2
 Fy  may :
or
v b2  2 gh
mv b2
up
R
bg
mb
2 ghg
 mg 
n b  mg 
nb

At the top of the loop, mgh 
R
FIG. P7.62
af
1
mv t2  mg 2 R
2
v t2  2 gh  4gR
209
210
Potential Energy
continued on next page
211
Chapter 7
 Fy  may :
n t  mg  
b
mb
2 ghg

 5mg
n t  mg 
nt
mv t2
R
m
2 gh  4 gR
R
g
R
Then the normal force at the bottom is larger by
n b  n t  mg 
P7.63
(a)
b g mb2 ghg 5mg 
m 2 gh
R
R
6mg .
Conservation of energy for the sled-rider-Earth
system, between A and C:
K i  U gi  K f  U gf
b
g me9.80 m s ja9.76 m f 12 mv  0
b2.5 m sg 2e9.80 m s ja9.76 m f  14.1 m s
1
m 2.5 m s
2
vC 
(b)
2
2
2
2
C
2
FIG. P7.63(a)
Incorporating the loss of mechanical energy during the portion of the motion in the water,
we have, for the entire motion between A and D (the rider’s stopping point),
b gb
1
80 kg 2.5 m s
2
Ki  U gi  fk d  K f  U gf :
g b80 kg ge9.80 m s ja9.76 mf f x  0  0
2
2
k
 f k d  7.90  10 3 J
(c)
fk 
and also a normal force of
n  mg  80 kg 9.80 m s 2  784 N
b ge
a158 N f  a784 N f
2
The magnitude of the water force is
(d)
7.90  10 3 J 7.90  10 3 N  m

 158 N
d
50 m
The water exerts a frictional force
2
j
 800 N

The angle of the slide is
  sin 1
9.76 m
 10.4
54.3 m

For forces perpendicular to the track at B,
 Fy  may :
FIG. P7.63(d)
n B  mg cos   0
b
ge
j
n B  80.0 kg 9.80 m s 2 cos 10.4  771 N

212
Potential Energy
continued on next page
(e)
 Fy  may :
n C  mg 
mv C2
r



b ge
j
b80.0 kg gb14.1 m sg

n C  80.0 kg 9.80 m s 2
2

20 m
n C  1.57  10 3 N up
FIG. P7.63(e)
The rider pays for the thrills of a giddy height at A, and a high speed and tremendous
splash at C. As a bonus, he gets the quick change in direction and magnitude among the
forces we found in parts (d), (e), and (c).
*P7.64
j a627 N fy
b ge
(a)
U g  mgy  64 kg 9.8 m s 2 y 
(b)
At the original height and at all heights above 65 m  25.8 m  39.2 m , the cord is
unstretched and U s  0 . Below 39.2 m, the cord extension x is given by x  39.2 m  y , so
the elastic energy is U s 
(c)
For y  39.2 m , U g  U s 
b
gb
1 2
1
kx 
81 N m 39.2 m  y
2
2
g.
2
a627 N fy
For y  39.2 m ,
a f
a f
e
 b
40.5 N m g
y b
2 550 N g
y  62 200 J
U g  U s  627 N y  40.5 N m 1 537 m 2  78.4 m y  y 2
2
j
Chapter 7
(d)
FIG. P7.64(d)
continued on next page
213
214
Potential Energy
(e)
At minimum height, the jumper has zero kinetic energy and the same total energy as at his
starting point. Ki  U i  K f  U f becomes
a fb
g b
g
627 N 65 m  40.5 N m y 2f  2 550 N y f  62 200 J
0  40.5 y 2f  2 550 y f  21 500
y f  10.0 m
(f)
the root 52.9 m is unphysical.
The total potential energy has a minimum, representing a stable equilibrium position. To
find it, we require
dU
 0:
dy
d
40.5 y 2  2 550 y  62 200  0  81y  2 550
dy
e
j
y  31.5 m
(g)
Maximum kinetic energy occurs at minimum potential energy. Between the takeoff point
and this location we have Ki  U i  K f  U f
0  40 800 J 
a f  2 550a31.5f 62 200
b g
1
2
64 kg v max
 40.5 31.5
2
F2b40 800  22 200gI 
G
J
H 64
K
2
12
v max
24.1 m s
ANSWERS TO EVEN PROBLEMS
P7.2
(a) 800 J; (b) 107 J; (c) 0
P7.4
(a) 1.11  10 J ; (b) 0.2
P7.6
(a) 19.8 m s ; (b) 294 J;
(c) 30.0 i  39.6 j m s
P7.18
9
e
j
P7.8
(a) 2.29 m/s; (b) 1.98 m/s
P7.10
(a) v B  5.94 m s , v C  7.67 m s ; (b) 147 J
P7.12
5.49 m/s
P7.14
(a) see the solution; (b) 35.0 J
P7.16
d
kx 2
x
2mg sin 
(a) 25.8 m; (b) 27.1 m s 2
Chapter 7
P7.20
(a) 0.403 m or  0.357 m ;
(b) see the solution
P7.22
(a) 21.0 m/s; (b) 16.1 m/s
P7.24
168 J
P7.26
(a) 24.5 m/s; (b) yes; (c) 206 m;
(d) unrealistic, see the solution
P7.28
3.92 kJ
P7.30
44.1 kW
P7.32
(a)
P7.34
e7  9x y ji  3x j
Ax 2 Bx 3
;

2
3
5 A 19B
5 A 19B
(b) U 
; K  


2
3
2
3
2
3
215
216
Potential Energy
P7.36
4.17  1010 J
P7.54
(a) 0.378 m; (b) 2.30 m/s; (c) 1.08 m
P7.38
(a) 1.67  1014 J; (b) at the center
P7.56
(a) see the solution; (b) 7.42 m/s
P7.40
(a) r  1.5 mm stable, 2.3 mm unstable,
3.2 mm stable, r   neutral;
(b) 5.6 J  E  1 J ;
P7.58
0.923 m/s
P7.60
2m
(c) 0.6 mm  r  3.6 mm ; (d) 2.6 J;
(e) 1.5 mm; (f) 4 J
P7.62
(a) 2.5R; (b) see the solution
P7.64
(a) 627 N y ;
(b) U s  0 for y  39.2 m and
1
2
U s  81 N m 39.2 m  y for
2
y  39.2 m ;
P7.42
33.4 kW  44.8 hp
P7.44
(a) 6.75 W m 2 ; (b) 6.64  103 W m 2 ;
a f
b
(c) see solution
gb
a f
U U  b
40.5 N m g
y
b
2 550 N g
y  62 200 J for y  39.2 m ;
(c) U g  U s  627 N y , for y  39.2 m
P7.46
48.2°
P7.48
(a) 0.588 J; (b) 0.588 J; (c) 2.42 m/s;
(d) U C  0.392 J, K C  0.196 J
2
g
P7.50
(a) 100 J; (b) 0.410 m; (c) 2.84 m/s;
(d) –9.80 mm; (e) 2.85 m/s
P7.52
(a) 3 x 2  4x  3 i ;
e
s
(d) see the solution; (e) 10.0 m;
(f) yes: stable equilibrium at 31.5 m;
(g) 24.1 m/s
j
(b) x  1.87 and –0.535;
(c) see the solution
CONTEXT 1 CONCLUSION SOLUTIONS TO PROBLEMS
*CC1.1
g
b
gb
1
1
mv 2  1 300 kg 22 m s
2
2
g
2
(a)
K
3.15  10 5 J
(b)
Energy delivered to battery  0.70 3.15  105 J  2.20  105 J
(c)
Energy returned to battery  0.85 2.20  105 J  1.87  105 J
(d)
Kinetic energy produced by motor  0.68 1.87  10 5 J  1.27  10 5 J
(e)
F2K I F2  1.27  10 JIJ 
v G J G
Hm K H 1 300 kg K
(f)
e
e
j
e
j
e
12
5
j
12
14.0 m s
useful energy output 1.27  10 5 J

 0.405  40.5%
total energy input
3.15  10 5 J
Chapter 7
continued on next page
217
218
Potential Energy
(g)
The original kinetic energy all becomes either final kinetic energy or internal energy:
3.15  10 5 J  1.27  10 5 J  Eint
Eint  1.87  10 5 J
*CC1.2
u sefu l ou tp u t en ergy
,
in p u t en ergy
33 MJ
33 MJ
33 MJ
, input 
0.07 
, input  110 MJ . Total
 471 MJ . Second part: 0.3 
input
input
0.07
66 MJ
input  471 MJ  110 MJ  581 MJ . For the hybrid car, 0.3 
, input  220 MJ .
input
(a)
For the conventional car in the first part of the trip, efficiency 
(b)
conventional car: efficiency 
66 MJ
 0.114  11.4% . This is much closer to 7% than to
input
30%. Hybrid car: efficiency  30.0%
ANSWERS TO EVEN CONTEXT 1 CONCLUSION PROBLEMS
CC1.2
(a) 581 MJ, 220 MJ; (b) 11.4%, 30.0%