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CHAPTER 5
Quick Quizzes
1.
(c), (a), (d), (b). The work in (c) is positive and of the largest possible value because the
angle between the force and the displacement is zero. The work done in (a) is zero
because the force is perpendicular to the displacement. In (d) and (b), negative work is
done by the applied force because in neither case is there a component of the force in the
direction of the displacement. Situation (b) is the most negative value because the angle
between the force and the displacement is 180°.
2.
All three balls have the same speed the moment they hit the ground because all start with
the same kinetic energy and undergo the same change in gravitational potential energy.
3.
(c).
4.
(c). The decrease in mechanical energy of the system is fkx. This is smaller than the value
on the horizontal surface for two reasons: (1) the force of kinetic friction fk is smaller
because the normal force is smaller, and (2) the displacement x is smaller because a
component of the gravitational force is pulling on the book in the direction opposite to its
velocity.
135
C H A P T E R
5
Problem Solutions
5.1
If the weights are to move at constant velocity, the net force on them must be zero. Thus,
the force exerted on the weights is upward, parallel to the displacement, with
magnitude 350 N. The work done by this force is
W   F cos   s   350 N  cos 0 2.00 m   700 J .
5.2
To lift the bucket at constant speed, the woman exerts an upward force whose
magnitude is F  mg   20.0 kg  9.80 m s2  196 N . The work done is W   F cos   s , so


the displacement is
s
W
6.00  103 J

 30.6 m .
F cos  196 N  cos 0




5.3
W   F cos   s   5.00  103 N cos 0 3.00  103 m  1.50  107 J  15.0 MJ .
5.4
The applied force makes an angle of 25° with the displacement of the cart. Thus, the
work done on the cart is
W   F cos   s   35 N  cos 25 50 m   1.6  103 J  1.6 kJ .
5.5


(a) The force of gravity is given by mg   5.00 kg  9.80 m s 2  49.0 N and is directed
downwards. The angle between the force of gravity and the direction of motion is
 = 90.0° - 30.0° = 60.0°, and so the work done by gravity is given as
W g   F cos   s   49.0 N  cos 60.0 2.50 m   61.3 J .
(b) The normal force exerted on the block by the incline is n  mg cos 30.0 , so the
friction force is
fk  k n   0.436 49.0 N  cos 30.0  18.5 N .
This force is directed opposite to the displacement (i.e.  = 180°), and the work it
does is
W f   fk cos   s   18.5 N  cos180  2.50 m    46.3 J .
(c) Since the normal force is perpendicular to the displacement;   90, cos  0 , and
the work done by the normal force is zero .
136
C H A P T E R
5.6
5
The total distance the scraper is moved over the surface of the tooth is
s  20  0.75 cm   15 cm  0.15 m . The friction force has magnitude
fk  k n   0.90 5.0 N   4.5 N . Hence, the force which must be applied in the direction
of the motion to overcome friction is F  4.5 N and the work done is
W   F cos   s   4.5 N  cos 0 0.15 m   0.68 J .
5.7
(a)
Fy  Fsin   n  mg  0
n  mg  F sin 
Fx  F cos   k n  0
n

F cos 
k
mg  F sin  
F cos 
k
 0.500 18.0 kg   9.80 m s 
k mg
F=
=
= 79.4 N
k sin   cos   0.500 sin 20.0  cos 20.0
2
(b) WF   F cos   s   79.4 N  cos 20.0 20.0 m   1.49  103 J  1.49 kJ
(c)
fk  Fcos   74.6 N
W f =  fk cos   s=  74.6 N  cos180 20.0 m    1.49  10 3 J   1.49 kJ
5.8
(a) WF   F cos   s  16.0 N  cos 25.0 2.20 m 
WF  31.9 J
(b) Wn   n cos 90 s  0
(c)
W g   mg cos 90 s  0
(d) Wnet  WF  Wn  Wg  31.9 J  0  0  31.9 J
137
C H A P T E R
5.9
5
(a) The work-energy theorem, Wnet  KEf  KEi , gives
5000 J 


1
2.50  103 kg v 2  0 , or
2
v = 2.00 m s .
(b) W   F cos   s   F cos 0  25.0 m   5000 J , so F  200 N .
5.10
1
Requiring that KEping pong  KEbowling with KE  mv 2 , we have
2


1
1
2
2.45  103 kg v 2  7.00 kg  3.00 m s  , giving v  160 m s .
2
2
5.11
w
700 N

 71.4 kg . The net upward force acting on the
g 9.80 m s 2
body is Fnet  2 355 N   700 N  10.0 N . The final upward velocity can then be calculated
from the work-energy theorem as
The person’s mass is m 
1
1
Wnet  KE f  KEi  mv 2  mvi2 ,
2
2
or
 Fnet cos   s  10.0 N  cos 00.250 m   2 71.4 kg  v 2  0
1
which gives
5.12
v = 0.265 m s upward .
(a) Wg  mg cos  s  mg cos  90.0     s


W g  10.0 kg  9.80 m s 2 cos110  5.00 m 
=  168 J
(b) W f   fk cos   s   k n  cos180 s
 0.400  98.0 N cos 20.0 cos180 5.00 m 
=  184 J
(c)
WF   F cos   s  100 N  cos 0 5.00 m   500 J
138
C H A P T E R
5
(d) KE  Wnet  Wg  W f  WF  148 J
(e)
1
1
KE  mv 2  mv i2
2
2
v
5.13
2  KE
2 148 J
2
 v i2 
 1.50 m s   5.64 m s
m
10.0 kg
(a) We use the work-energy theorem to find the work.
1
1
1
2
2
W  KE  mv 2  mv i2  0  70 kg  4.0 m s    5.6  10 J .
2
2
2
(b) W   F cos   s   fk cos180 s    k n  s    k mg  s ,
so
5.14
s


5.6  102 J
W

 1.2 m .
k mg
 0.70 70 kg  9.80 m s 2


At the top of the arc, v y  0 , and v x  v ix  v i cos 30.0  34.6 m s .
Therefore v 2  v x2  v y2   34.6 m s  , and
2
1
1
2
KE  mv 2   0.150 kg  34.6 m s   90.0 J .
2
2
5.15
(a) The final kinetic energy of the bullet is


1
1
2
KE f  mv 2  2.0  103 kg  300 m s   90 J .
2
2
(b) We know that
Thus, F 
5.16
(a)


W  KE , and also W  F cos  s .
KE
90 J  0
2

 1.8  10 N
s cos   0.50 m  cos 0
1
1
2
KEA  mv A2   0.60 kg  2.0 m s   1.2 J
2
2
139
C H A P T E R
5
2  KEB 
2  7.5 J
1

 5.0 m s
(b) KEB  mv B2 , so v B 
m
0.60 kg
2
(c) Wnet  KE  KEB  KEA   7.5  1.2 J  6.3 J
5.17
Wnet   Froad cos 1  s   Fresist cos 2  s  1000 N  cos 0s   950 N  cos180s
Wnet  1000 N-950 N  20 m   1.0  103 J
1
Also, Wnet  KE f  KEi  mv 2  0 , so
2


2 1.0  103 J
2Wnet
v

 1.0 m s
m
2000 kg
5.18
The initial kinetic energy of the sled is
1
1
2
KEi  mvi2  10 kg  2.0 m s   20 J ,
2
2
and the friction force is fk  k n  k mg   0.10 98 N   9.8 N .
Wnet   fk cos180 s  KE f  KEi , so s 
5.15
0  KEi
20 J

 2.0 m
fk cos180 9.8 N
(a) The final kinetic energy of the bullet is


1
1
2
KE f  mv 2  2.0  103 kg  300 m s   90 J .
2
2
(b) We know that
Thus, F 
5.16
(a)


W  KE , and also W  F cos  s .
KE
90 J  0
2

 1.8  10 N
s cos   0.50 m  cos 0
1
1
2
KEA  mv A2   0.60 kg  2.0 m s   1.2 J
2
2
140
C H A P T E R
5
2  KEB 
2  7.5 J
1

 5.0 m s
(b) KEB  mv B2 , so v B 
m
0.60 kg
2
(c) Wnet  KE  KEB  KEA   7.5  1.2 J  6.3 J
5.17
Wnet   Froad cos 1  s   Fresist cos 2  s  1000 N  cos 0s   950 N  cos180s
Wnet  1000 N-950 N  20 m   1.0  103 J
1
Also, Wnet  KE f  KEi  mv 2  0 , so
2


2 1.0  103 J
2Wnet
v

 1.0 m s
m
2000 kg
5.18
The initial kinetic energy of the sled is
1
1
2
KEi  mvi2  10 kg  2.0 m s   20 J ,
2
2
and the friction force is fk  k n  k mg   0.10 98 N   9.8 N .
Wnet   fk cos180 s  KE f  KEi , so s 
5.19
0  KEi
20 J

 2.0 m
fk cos180 9.8 N


The weight of the blood is mg   0.50 kg  9.80 m s 2  4.9 N .
(a) The displacement from the feet to the heart is y  1.3 m so the potential energy,
using the feet as the reference level, is
PEg  mgy   4.9 N 1.3 m   6.4 J .
(b) The displacement from the head to the heart is y  1.3  1.8 m  0.50 m . Thus,
when the head is used as the reference level, the potential energy is
PEg  mgy   4.9 N  0.50 m    2.5 J .
141
C H A P T E R
5.20
5
(a) Relative to the ceiling, y = -1.00 m.


Thus, PEg  mgy   2.00 kg  9.80 m s2  1.00 m    19.6 J .
(b) Relative to the floor, y = 2.00 m, so


PEg  mgy   2.00 kg  9.80 m s 2  2.00 m   39.2 J .
(c) Relative to the height of the ball, y = 0, and PEg  0 .
5.21
The work the muscle must do against the force of gravity to raise the arm is




Wmuscle  W g   mgy i  mgy f  mg y f  y i  mgh  mgL 1  cos 30 ,
or
5.22


Wmuscle  7.0 kg  9.80 m s2  0.21 m 1  cos 30  1.9 J .
When the legs are straightened, the spring will be stretched by
x   0.720 m 1  sin 40.0   0.740 m 1  sin 25.0  0.684 m .
The work the legs must do to stretch the spring this amount is
1
1
2
W  PEs  kx 2   250 N m   0.684 m   58.6 J .
2
2
5.23
The equivalent spring constant for the muscle is
k
F
105 N

 5.25  103 N m .
x 2.00  10-2 m
The work done in stretching this by 2.00 cm is


1
1
W  PEs  kx 2  5.25  103 N m 2.00  10-2 m
2
2
142

2
 1.05 J .
C H A P T E R
5.24
5
Let m be the mass of the ball, R the radius of the circle, and F the 30.0 N force. With y = 0
at the bottom of the circle, Wnc   KE  PE f   KE  PEi yields
 F cos 0  R   2 mv 2f  0   2 mv i2  mg  2R  ,
1
or
vf 
Thus, v f 
1
2F   R 
 v i2  4 gR .
m
2  30.0 N    0.600 m 
2
 15.0 m s   4 9.80 m s 2  0.600 m 
0.250 kg


giving v f  26.5 m s
5.25
(a) When the rope is horizontal, the swing is 2.0 m above the level of the bottom of the
circular arc, and PEg  mgy   40 N  2.0 m   80 J .
(b) When the rope makes a 30° angle with the vertical, the vertical distance from the
swing to the lowest level in the circular arc is
y  L  L cos 30  L1  cos 30 and the potential energy is given by
PEg  mgy  mgL 1  cos 30   40 N  2.0 m 1  cos 30  11 J .
(c) At the bottom of the circular arc, y  0 . Hence, PEg  0 .
5.26
Using conservation of mechanical energy, we have
1 2
1
mv f  mgy f  mvi2  0 ,
2
2
or
yf 
v i2  v 2f
2g
10 m s   1.0 m s 

2

2 9.80 m s 2
2

143
 5.1 m .
C H A P T E R
5.27
5
Since no non-conservative forces do work, we use conservation of mechanical energy,
with the zero of potential energy selected at the level of the base of the hill. Then,
1 2
1
mv f  mgy f  mv i2  mgy i with y f  0 yields
2
2
yi 
v 2f  v i2
2g
 3.00 m s 

2

0
2 9.80 m s 2

 0.459 m .
Note that this result is independent of the mass of the child and sled.
5.28
(a) We take the zero of potential energy at the level of point B, and use conservation of
1
1
mechanical energy to obtain mv B2  0  mv A2  mgy A , or
2
2


v B  v A2  2gy A  0  2 9.80 m s 2  5.00 m   9.90 m s .
(b) At point C, with the starting point at A, we again use conservation of mechanical
1
1
energy. This gives mvC2  mgyC  mv A2  mgy A , and yields
2
2


vC  v A2  2g  y A  yC   0  2 9.80 m s 2  3.00 m   7.67 m s .
5.29
Using conservation of mechanical energy, starting when point A is directly over the bar
and ending when it is directly below the bar, gives
1 2
1
mv f  mgy f  mv i2  mgy i , or
2
2




v f  v i2  2 g y i  y f  0  2 9.80 m s 2  2.40 m   6.86 m s
144
C H A P T E R
5.30
5
(a) From conservation of mechanical energy,
1 2
1
mv B  mgy B  mv A2  mgy A , or
2
2
v B  v A2  2 g  y A  y B 


 0  2 9.80 m s 2 1.80 m   5.94 m s .
Similarly,
v C  v A2  2 g  y A  y B   0  2 g  5.00 m  2.00 m   7.67 m s .
(b) Wg
5.31

A C
    PE 
 PEg
A
g
C
 mg  y A  yC    49.0 N  3.00 m   147 J
(a) We choose the zero of potential energy at the level of the bottom of the arc. The
initial height of Tarzan above this level is
yi   30.0 m 1  cos 37.0  6.04 m .
Then, using conservation of mechanical energy, we find
1 2
1
mv f  0  mvi2  mgyi , or
2
2


v f  vi2  2gyi  0  2 9.80 m s 2  6.04 m   10.9 m s .
(b) In this case, conservation of mechanical energy yields
v f  v i2  2 gy i 
 4.00 m s 2  2  9.80 m s 2  6.04 m  
145
11.6 m s .
C H A P T E R
5.32
5
Realize that all three masses have identical speeds at each point in the motion and that
vi  0 . Then, conservation of mechanical energy gives
KEf  PEi  PEf , or






1
 m1  m2  m3  v 2f  m1 y1i  y1 f  m2 y 2i  y 2 f  m3 y 3i  y 3 f  g
2
Thus,

1
 30.0 v 2f   5.00 4.00 m   10.00  15.0 4.00 m  9.80 m s 2
2
v f  5.11 m s .
yielding
5.33

(a) Use conservation of mechanical energy from when the projectile is at rest within the
gun until it reaches maximum height.

Then, KE  PEg  PEs
   KE  PE

 PEs becomes
g
f
i
1
0  mgy max  0  0  0  kxi2 ,
2
or



3
2
2mgy max 2 20.0  10 kg 9.80 m s  20.0 m 
k

 544 N m .
x i2
 0.120 m  2
(b) This time, we use conservation of mechanical energy from when the projectile is at
rest within the gun until it reaches the equilibrium position of the spring. This gives

 
KE f  PEg  PEs  PEg  PEs

i

f
1


  mgx i  kx i2    0  0


2
 k
v 2f    x i2  2 gx i
 m
 544 N m 

 0.120 m  2  2 9.80 m s 2  0.120 m 
3

20.0

10
kg




yielding v f  19.7 m s
146


C H A P T E R
5.34
5
At maximum height, v y =0 and v x =v ix   40 m s  cos 60  20 m s .
Thus, v f  v x2  v y2  20 m s . Choosing PEg  0 at the level of the launch point,
conservation of mechanical energy gives PEf  KEi  KEf , and the maximum height
reached is
yf 
5.35
v i2  v 2f
2g
 40 m s    20 m s 

2

2 9.80 m s 2
2

 61 m .
Choose PEg  0 at the level of the release point and use conservation of mechanical
energy from release until the block reaches maximum height. Then,

KEf  KEi  0 and we have PEg  PEs

g
f

 PEs , or
i
1
mgy max  0  0  kxi2 which yields
2
y max
5.36
   PE


5.00  103 N m  0.100 m 
kx i2


 10.2 m .
2mg
2  0.250 kg  9.80 m s 2

2

(a) Choose PEg  0 at the level of point B. Between A and B, we can use conservation
 
of mechanical energy, KEB  PEg
B
 
 KEA  PEg
A
, which becomes
1 2
mv B  0  0  mgy A or
2


v B  2 g y A  2 9.80 m s 2  5.00 m   9.90 m s .
(b) Again, choose PEg  0 at the level of point B. Between points B and C, we use the
work-kinetic energy theorem in the form

   KE  PE 
1
 0  mgyC  mv B2  0 to find
2
1
2
Wnc   0.400 kg  9.80 m s 2  2.00 m   0.400 kg  9.90 m s   11.8 J .
2
Wnc  KE  PEg
g
C

B

Thus, 11.8 J of energy is spent overcoming friction between B and C.
147
C H A P T E R
5.37
5
The work done by the non-conservative resistance force is


Wnc   f cos   s   2.7  103 N cos180  0.20 m   5.4  10 4 J
We use the work-kinetic theorem in the form

Wnc  KE  PEg
Thus,
g
f
i

1 2
1
mv f  Wnc  mvi2  mg y i  y f
2
2
vf 

5.38
   KE  PE  , or KE



 2.5  10
2 5.4  104 J
4.2  10 kg
-3
   
 Wnc  KEi  PEg  PEg
i
f
.
 or

2Wnc
 v i2  2g y i  y f
m
f
2



m s  2 9.80 m s 2  0.20 m   1.9 m s
2
The friction force does work W friction on the gurney during the 20.7 m horizontal
displacement at constant speed. The attendant does 2000 J of work on the gurney during
this motion. Applying the work-kinetic energy theorem gives
Wnc  W friction  2000 J   KE  PE f   KE  PEi  0 , so W friction  2000 J .
Also, W friction   f cos180  s   f s . Therefore, the friction force is given by
f
W friction
s

  2000 J
 96.6 N .
20.7 m
Now consider the displacement s  0.48 m that occurs after the gurney is released with
an initial speed of v i  0.88 m s . Applying the work-kinetic energy theorem to this part
of the trip gives

Wnc   f cos180 s  KE  PEg
or
   KE  PE  ,
g
f
i
1

 f s   0  0   mv i2  0 which gives the total mass of the gurney plus patient
2

as m  2 f s v i2 , and the total weight as
 2  96.6 N  0.48 m  
 2 f s 
 9.80 m s 2  1.2  103 N  1.2 kN
w  mg   2  g  
2
 vi 
  0.88 m s 


148

C H A P T E R
5.39
5
We shall take PEg  0 at the lowest level reached by the diver under the water. The
diver falls a total of 15 m, but the non-conservative force due to water resistance acts
only during the last 5.0 m of fall. The work-kinetic energy theorem then gives

Wnc  KE  PEg
or
   KE  PE  ,
g
f
i
 F cos180  5.0 m   0  0  0  70 kg   9.80 m s  15 m   .
2
This gives the average resistance force as F  2.1  103 N  2.1 kN .
5.40
    PE 
Since the plane is in level flight, PEg
g
f
i
and the work-kinetic energy theorem
reduces to Wnc  Wthrust  Wresistance  KEf  KEi , or
 Fcos 0 s   f cos180 s  2 mv 2f  2 mvi2 .
1
1
This gives
vf  v 
2
i
5.41
2 F  f  s
m
2  7.5  4.0  104 N   500 m 
  60 m s  
 77 m s
1.5  104 kg
2
Choose PEg  0 at the level of the bottom of the driveway.

Then Wnc  KE  PEg
   KE  PE 
g
f
i
becomes
 f cos180 s   2 mv 2f  0  0  mg  s sin 20  .
1


Solving for the final speed gives v f 
or

 2gs sin 20 

v f  2 9.80 m s 2  5.0 m  sin 20 
149

2f s
, or
m

2 4.0  103 N  5.0 m 
2.10  10 kg
3
 3.8 m s .
C H A P T E R
5.42
5
(a) Choose PEg  0 at the level of the bottom of the arc. The child’s initial vertical
displacement from this level is
yi   2.00 m 1  cos 30.0  0.268 m .
In the absence of friction, we use conservation of mechanical energy as
 KE  PE    KE  PE  , or 12 mv
g
g
f
i

2
f
 0  0  mgy i , which gives

v f  2gyi  2 9.80 m s 2  0.268 m   2.29 m s .
(b) With a non-conservative force present, we use

Wnc  KE  PEg
   KE  PE    12 mv
g
f
i
2
f

 0   0  mgy i  , or

 v 2f

Wnc  m   gy i 
 2

  2.00 m s  2

  25.0 kg  
 9.80 m s 2  0.268 m    15.6 J
2




Thus, 15.6 J of energy is spent overcoming friction.
5.43

(a) We use conservation of mechanical energy, KE  PEg
   KE  PE  , for the trip
g
f
down the frictionless ramp. With vi  0 , this reduces to



v f  2 g y i  y f  2 9.80 m s 2
  3.00 m  sin 30.0 
5.42 m s .
(b) Using the work-kinetic energy theorem,

Wnc  KE  PEg
   KE  PE  ,
g
f
i
for the trip across the rough floor gives
 fk cos180 s   0  0   2 mv i2  0 , or
1
150
1
fk s  k  mg  s  mv i2 .
2
i
C H A P T E R
5
Thus, the coefficient of kinetic friction is
 5.42 m s 
v2
k  i 
 0.300 .
2 g s 2 9.80 m s 2  5.00 m 
2




(c) All of the initial mechanical energy, KE  PEg  0  mgyi , is spent overcoming
i
friction on the rough floor. Therefore, the energy “lost” is


mgyi  10.0 m  9.80 m s 2  3.00 m  sin 30.0  147 J .
5.44
Choose PEg  0 at water level and

use KE  PEg
   KE  PE 
f
g
i
for
the trip down the curved slide.
This gives
1 2
 h
mv  mg    0  mgh , so the
 5
2
speed of the child as she leaves
the end of the slide is v 
2 g  4 h 5 .
The vertical component of this launch velocity is
 4h 
v iy  v sin   sin  2 g   .
 5
At the top of the arc, v y  0 . Thus, v y2  v iy2  2ay  y  gives the maximum height the child
reaches during the airborne trip as
h
  4h  

0  sin 2  2 g     2   g   y max  

5
  5 
This may be solved for y max to yield y max 
151


h
4sin 2   1 .
5
C H A P T E R
5.45
5
Choose PEg  0 at the level of the base of the hill and let x represent the distance the
skier moves along the horizontal portion before coming to rest. The normal force
exerted on the skier by the snow while on the hill is n1  mg cos10.5 and, while on the
horizontal portion, n 2  mg .
Consider the entire trip, starting from rest at the top of the hill until the skier comes to
rest on the horizontal portion. The work done by friction forces is
Wnc   fk 1 cos180  200 m    fk  2 cos180 x
  k  mg cos10.5  200 m   k  mg  x

Applying Wnc  KE  PEg
   KE  PE 
g
f
i
to this complete trip gives
 k  mg cos10.5  200 m   k  mg  x   0  0  0  mg  200 m  sin 10.5 ,
 sin 10.5

x 
 cos10.5  200 m  . If k  0.0750 , then x = 289 m .
 k

or
5.46
The normal force exerted on the sled by the track is n  mg cos  and the friction force is
fk  k n  k mg cos  .
If s is the distance measured along the incline that the sled travels, applying

Wnc  KE  PEg
   KE  PE 
f
g
i
to the entire trip gives
1
 k mg cos   cos180 s  0  mg s  sin      mv i2  0  ,
2

 4.0 m s 
v i2
s

 1.5 m .
2
2 g  sin   k cos   2 9.80 m s  sin 20  0.20cos 20
2
or


152
C H A P T E R
5.47
5

(a) Consider the entire trip and apply Wnc  KE  PEg
   KE  PE 
g
f
i
to obtain
 f1 cos180 d1   f2 cos180 d2   2 mv 2f  0   0  mgy i  , or
1
vf 
fd fd 

2  g yi  1 1 2 2 


m

 50.0 N  800 m    3600 N  200 m  
 2  9.80 m s 2 1000 m  

80.0 kg



which yields v f  24.5 m s .
(b)
Yes , this is too fast for safety.

(c) Again, apply Wnc  KE  PEg
   KE  PE  , now with d
g
f
2
i
considered to be a
variable, d1  1000 m  d2 , and v f  5.00 m s . This gives
 f1 cos180 1000 m  d2    f2 cos180 d2   2 mv 2f  0   0  mgy i  ,
1
1
which reduces to  1000 m  f1  f1 d2  f2 d2  mv 2f  mgy i . Therefore,
2
d2 

 mg yi  1000 m  f1  12 mv 2f
f2  f1
784 N 1000 m   1000 m  50.0 N    80.0 kg  5.00 m s 
3600 N  50.0 N
1
2
2
 206 m
(d) In reality. 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.
5.48
(a) Wnc  KE  PE , but KE  0 because the speed is constant. The skier rises a
vertical distance of y   60 m  sin30  30 m . Thus,


Wnc  70 kg  9.80 m s2  30 m   2.06  104 J 21 kJ .
153
C H A P T E R
5
(b) The time to travel 60 m at a constant speed of 2.0 m s is 30 s. Thus, the required
power input is

5.49
Wnc 2.06  104 J
 1 hp 

  686 W  
 0.92 hp .
 746 W 
t
30 s
(a) The work done by the student is


W  PEg  mg  y    50.0 kg  9.80 m s 2  5.00 m   2.45  103 J .
The time to do this work if she is to match the power output of a 200 W light bulb is
W 2.45  103 J
t

 12.3 s , so the required average speed is

200 W
v
y 5.00 m

 0.408 m s .
t
12.3 s
(b) The work done was found in part (a) as W  2.45  103 J= 2.45 kJ .
5.50
Let N be the number of steps taken in time t. We determine the number of steps per
unit time by
Power =
or
work done  work per step per unit mass   mass   # steps 
,

t
t

J step 
 N 
, giving
70 W   0.60
60 kg  


 t 
kg 

N
= 1.9 steps s .
t
The running speed is then
v
5.51
step  
x  N 
m 


distance traveled per step    1.9
1.5
 2.9 m s





t  t 
s 
step 
The power output is given by

or
energy spent  m g h

 1.2  106 kg s 9.80 m s 2  50 m  ,
t
t


 5.9  108 W= 5.9  102 MW .
154

C H A P T E R
5.52
5
(a) We use the work-kinetic energy theorem with KEi  0 and PE  0 to find
1
2
1

W  KE  PE   mv 2  0   0  1500 kg 10.0 m s 
2

2
W  7.50  104 J  75.0 kJ .
(b) The average power input is given by

W 7.50  104 J  1 hp 


  33.5 hp .
t
3.00 s  746 W 
(c) The acceleration of the car is
a
v 10.0 m s  0

 3.33 m s 2 .
t
3.00 s
Thus, the net force acting on the car is



F  ma  1.50  103 kg 3.33 m s 2  5.00  103 N .
At t = 2.00 s, the instantaneous velocity is


v  vi  at  0  3.33 m s 2  2.00 s   6.66 m s
and the instantaneous power input is
 1 hp  44.7 hp
 Fv  5.00  103 N  6.66 m s  

 746 W 

5.53

(a) The acceleration of the car is a 
v f  vi
t

18.0 m s  0
 1.50 m s 2 . Thus, the
12.0 s
constant forward force due to the engine is found from

.

F  Fengine  Fair  ma as

Fengine  Fair  ma  400 N  1.50  103 kg 1.50 m s2  2.65  103 N
The average velocity of the car during this interval is v 
v f  vi
average power input from the engine during this time is
 1 hp 
 Fengine v  2.65  103 N  9.00 m s  
 32.0 hp .
 746 W 


155
2
 9.00 m s , so the
C H A P T E R
5
(b) At t  12.0 s , the instantaneous velocity of the car is v  18.0 m s and the
instantaneous power input from the engine is
 1 hp 
 Fengine v  2.65  103 N 18.0 m s  
 63.9 hp .
 746 W 

5.54

(a) The acceleration of the elevator during the first 3.00 s is
a
v f  vi
t

1.75 m s  0
 0.583 m s 2 ,
3.00 s
so Fnet  Fmotor  mg  ma gives the force exerted by the motor as
Fmotor  m  a  g   650 kg   0.583  9.80 m s 2   6.75  103 N .
The average velocity of the elevator during this interval is v 
v f  vi
2
 0.875 m s ,
so the average power input from the motor during this time is
 1 hp 
 Fmotor v  6.75  103 N  0.875 m s  
 7.92 hp .
 746 W 


(b) When the elevator moves upward with a constant speed of v  1.75 m s , the
upward force exerted by the motor is Fmotor  mg and the instantaneous power input
from the motor is
 1 hp 
  mg  v   650 kg  9.80 m s 2 1.75 m s  
 14.9 hp .
 746 W 

5.55

The work done on the particle by the force F as
the particle moves from x  xi to x  x f is the area
under the curve from xi to x f .
(a) For x  0 to x  8.00 m ,
1
W  area of triangle ABC  AC  altitude
2
W08 
1
 8.00 m  6.00 N   24.0 J
2
156
C H A P T E R
5
(b) For x  8.00 m to x  10.0 m ,
1
W810  area of triangle CDE  CE  altitude
2
=
1
 2.00 m   3.00 N    3.00 J
2
(c) W010  W08  W810  24.0 J    3.00 J  21.0 J
5.56
W equals the area under the Force-Displacement Curve
(a) For the region 0  x  5.00 m ,
W
 3.00 N  5.00 m 
2
 7.50 J
(b) For the region 5.00 m  x  10.0 m ,
W   3.00 N  5.00 m   15.0 J
(c) For the region 10.0 m  x  15.0 m , W 
 3.00 N  5.00 m 
2
 7.50 J
(d) For the region 0  x  15.0 m , W   7.50  15.0  7.50 J  30.0 J
5.57
(a)
Fx   8x  16 N
(b) Wnet 
5.58
  2.00 m  16.0 N 
2

1.00 m  8.00 N 
2
  12.0 J
From the work-kinetic energy theorem, Wnet  KEf  KEi , we have
F
applied

1
cos 0 s  mv 2f  0 .
2
The mass of the cart is m 
vf 
2 Fapplied s
m

w
 10.0 kg , and the final speed of the cart is
g
2  40.0 N  12.0 m 
10.0 kg
157
 9.80 m s .
C H A P T E R
5.59
5
(a) The equivalent spring constant of the bow is given by F  kx as
k
Ff

xf
230 N
= 575 N m .
0.400 m
(b) From the work-kinetic energy theorem applied to this situation,

Wnc  KE  PEg  PEs
   KE  PE
g
f
 PEs

f
1


  0  0  kx 2f    0  0  0 .


2
The work done pulling the bow is then
1
1
2
Wnc  kx 2f   575 N m   0.400 m   46.0 J .
2
2
5.60
Choose PEg  0 at the level where the block comes to rest against the spring. Then, in
the absence of work done by non-conservative forces, the conservation of mechanical
energy gives
 KE  PE
g
or
 PEs
   KE  PE
g
f

 PEs ,
i
1
0  0  kx 2f  0  mg L sin   0 . Thus,
2


2 12.0 kg  9.80 m s 2  3.00 m  sin 35.0
2 mg L sin 
xf 

 0.116 m .
k
3.00  10 4 N m
5.61
(a) From v 2  v i2  2ay  y  , we find the speed just before touching the ground as


v  0  2 9.80 m s 2 1.0 m   4.4 m s .
(b) Choose PEg  0 at the level where the feet come to rest. Then

Wnc  KE  PEg
   KE  PE 
g
f
i
 F cos180 s   0  0   12 mv
becomes
2
i

 mg s

75 kg  4.4 m s   75 kg 9.80 m s 2  1.5  105 N
mv i2
 mg 
or F 


2s
2 5.0  103 m
2


158


C H A P T E R
5.62
5
From the work-kinetic energy theorem,

Wnc  KE  PEg  PEs
   KE  PE
g
f

 PEs ,
i
1
1
 

we have  fk cos180 s   mv 2f  0  0   0  0  kx i2  , or
2
 

2
8.0 N m   5.0  102 m 
kx i2  2 fk s
vf 

m
5.63
2
5.3  10
 2  0.032 N  0.15 m 
3
kg
 1.4 m s
(a) The two masses will pass when both are at y f  2.00 m above the floor. From

conservation of energy, KE  PEg  PEs
   KE  PE
g
f

 PEs ,
i
1
 m1  m2  v 2f  m1  m2  gy f  0  0  m1 gy1i  0 , or
2
vf 

2 m1 g y1i
 2gyf
m1  m2


2  5.00 kg  9.80 m s 2  4.00 m 
8.00 kg


 2 9.80 m s 2  2.00 m  .
This yields the passing speed as v f  3.13 m s .
(b) When m1  5.00 kg reaches the floor, m2  3.00 kg is y 2 f  4.00 m above the floor.

Thus, KE  PEg  PEs
   KE  PE
f
g
 PEs

i
becomes
1
 m1  m2  v 2f  m2 g y 2 f  0  0  m1 gy1i  0 , or v f 
2

2 g m1 y1i  m2 y 2 f

m1  m2
Thus,
vf 


2 9.80 m s 2  5.00 kg   4.00 m    3.00 kg   4.00 m  
 4.43 m s
8.00 kg
159
C H A P T E R
5
(c) When the 5.00-kg hits the floor, the string goes slack and the 3.00-kg becomes a
projectile launched straight upward with initial speed viy  4.43 m s . At the top of
its arc, v y2  v iy2  2ay  y  gives
y 
5.64
v y2  v iy2
2 ay

0   4.43 m s 

2 9.80 m s 2
2

 1.00 m .
The normal force the incline exerts on block A is n A   mA g  cos 37 , and the friction force
is fk  k nA  k mA g cos 37 . The vertical distance block A rises is
y A   20 m  sin 37  12 m , while the vertical displacement of block B is y B  20 m .
We find the common final speed of the two blocks by use of

Wnc  KE  PEg
   KE  PE   KE  PE .
g
f
g
i
1

This gives   k mA g cos 37 s    m A  mB  v 2f  0   m A g  y A   mB g  y B  
2


or
v 2f 

2 g mB  y B   m A  y A    k m A cos 37 s 
m A  mB


2 9.80 m s 2  100 kg   20 m    50 kg  12 m   0.25  50 kg   20 m  cos 37
150 kg
which yields v 2f  157 m 2 s 2 .
The change in the kinetic energy of block A is then


1
1
KEA  mA v 2f  0   50 kg  157 m 2 s 2  3.9  103 J  3.9 kJ .
2
2
5.65
Since the Marine moves at constant speed, the upward force the rope exerts on him
must equal his weight, or F  700 N . The constant speed at which he moves up the rope
y 10.0 m
is v 

 1.25 m s , so the constant power output is
t 8.00 s
  Fv   700 N  1.25 m s   875 W .
160
C H A P T E R
5.66
5
When 1 pound (454 grams) of fat is metabolized, the energy released is
E   454 g  9.00 kcal g   4.09  103 kcal . Of this, 20.0% goes into mechanical energy
(climbing stairs in this case). Thus, the mechanical energy generated by metabolizing 1
pound of fat is


Em   0.200 4.09  103 kcal  817 kcal .
Each time the student climbs the stairs, she raises her body a vertical distance of
y   80 steps  0.150 m step   12.0 m . The mechanical energy required to do this is
PEg  mg  y  , or
 1 kcal 
PEg   50.0 kg  9.80 m s 2 12.0 m   5.88  10 3 J 
 1.40 kcal .
 4186 J




(a) The number of times the student must climb the stairs to metabolize 1 pound of fat
E
817 kcal
 582 trips .
is N  m 
PEg 1.40 kcal trip
It would be more practical for her to reduce food intake.
(b) The useful work done each time the student climbs the stairs is W  PEg  5.88  103 J .
Since this is accomplished in 65.0 s, the average power output is

5.67
W 5.88  103 J
 1 hp 

 90.5 W =  90.5 W 
 0.121 hp .
 746 W 
t
65.0 s
 4186 J
 9.21  105 J of energy while going
(a) The person walking uses Ew   220 kcal  
 1 kcal 
3.00 miles. The quantity of gasoline which could furnish this much energy is
9.21  105 J
V1 
 7.08  103 gal . This means that the walker’s fuel economy in
8
1.30  10 J gal
equivalent miles per gallon is
fuel economy 
3.00 mi
 423 mi gal .
7.08  10-3 gal
161
C H A P T E R
5
(b) In 1 hour, the bicyclist travels 10.0 miles and uses
 4186 J
EB   400 kcal  
 1.67  106 J ,
 1 kcal 
which is equal to the energy available in
1.67  106 J
 1.29  102 gal of gasoline. Thus, the equivalent fuel economy for
1.30  108 J gal
10.0 mi
the bicyclist is
 776 mi gal .
1.29  102 gal
V2 
5.68
We use Wnet  KEf  KEi to find the average force the hand exerts on the baseball.
m v2
1
Wnet  F cos180 s  0  mball v i2 , so F  ball i .
2s
2


The force the ball exerts on the hand is in the opposite direction but has the same
magnitude.
(a) If s  2.0 cm  2.0  102 m ,
 0.15 kg  25 m s 
F

2 2.0  102 m
2
 2.3  103 N= 2.3 kN .

(b) Similarly, if s  10 cm  0.10 m , we find F  4.7  102 N .
5.69

(a) Use conservation of mechanical energy, KE  PEg
   KE  PE  , from the start to
f
g
i
the end of the track to find the speed of the skier as he leaves the track. This gives
1 2
mv  mgy f  0  mgyi , or
2
v


2 g yi  y f 


2 9.80 m s 2  40.0 m   28.0 m s .
162
C H A P T E R
5
(b) At the top of the parabolic arc the skier follows after leaving the track, v y  0 and
v x   28.0 m s  cos 45.0  19.8 m s . Thus, v top  v x2  v y2  19.8 m s . Applying
conservation of mechanical energy from the end of the track to the top of the arc
1
1
2
2
gives m 19.8 m s   mg y max  m  28.0 m s   mg 10.0 m  , or
2
2
 28.0 m s   19.8 m s 
 10.0 m 
2
y max

2 9.80 m s 2

2
 30.0 m .
1
(c) Using y  viy t  ay t 2 for the flight from the end of the track to the ground gives
2
10.0 m   28.0 m s  sin 45.0 t 


1
9.80 m s 2 t 2
2
The positive solution of this equation gives the total time of flight as t  4.49 s .
During this time, the skier has a horizontal displacement of
x  vix t   28.0 m s cos 45.0  4.49 s  89.0 m .
5.70
First, determine the magnitude of the applied force by
considering a free-body diagram of the block. Since the block
moves with constant velocity, Fx  Fy  0 .
From Fx  0 , we see that n  F cos 30 .
Thus, fk  k n  k Fcos 30 , and Fy  0 becomes
F sin 30  mg  k F cos 30 , or
F


 5.0 kg  9.80 m s 2
mg

 2.0  102 N .
sin 30  k cos 30 sin 30   0.30 cos 30
(a) The applied force makes a 60° angle with the displacement up the wall. Therefore,


WF   F cos 60 s   2.0  102 N cos 60  3.0 m   3.1  102 J .
(b) W g   mg cos180 s   49 N  1.0 3.0 m    1.5  102 J .
(c)
Wn   n cos 90 s  0 .
163
C H A P T E R
5
(d) PEg  mg  y    49 N  3.0 m   1.5  102 J .
5.71
The force constant of the spring is k  1.20 N cm  120 N m . If the spring is initially
compressed a distance x i , the vertical distance the ball rises as the spring returns to the
equilibrium position is
y f  xi sin 10.0

In the absence of friction, we apply KE  PEg  PEs
   KE  PE
f
g
 PEs

i
from the release
of the plunger to when the spring has returned to the equilibrium position and obtain
1 2
1
mv f  mg  xi sin 10.0  0  0  0  kxi2 , or
2
2
vf 

This yields
5.72
kx i2
 2 g x i sin 10.0
m
120 N m   5.00  102 m 
0.100 kg
2



 2 9.80 m s 2 5.00  102 m sin 10.0
v f  1.68 m s .
If a projectile is launched, in the absence of air resistance, with speed v i at angle 
above the horizontal, the time required to return to the original level is found from
g
2 v sin 
1
. The range is the
y  viy t  ay t 2 as 0   v i sin   t  t 2 , which gives t  i
2
2
g
horizontal displacement occurring in this time.
 2 v sin   v i2  2sin  cos   v i2 sin  2 
Thus, R  v ix t   v i cos    i
.


g 
g
g

Maximum range occurs when   45 , giving v i2  g Rmax . The minimum kinetic energy
required to reach a given maximum range is
1
1
KE  mvi2  mg Rmax
2
2
164
C H A P T E R
5
(a) The minimum kinetic energy needed in the record throw of each object is
KE 
Javelin:


1
2
0.80 kg  9.80 m s 2  89 m   3.5  10 J

2




Discus: KE 
1
2
2.0 kg  9.80 m s 2  69 m   6.8  10 J

2
KE 
1
7.2 kg  9.80 m s 2  21 m   7.4  102 J
2
Shot:
(b) The average force exerted on an object during launch, when it starts from rest and
is given the kinetic energy found above, is computed from Wnet  Fs  KE as
F
KE  0
. Thus, the required force for each object is
s
F
Javelin:
Discus: F 
6.8  102 J
2
 3.4  10 N
2.00 m
F
7.4  102 J
2
 3.7  10 N
2.00 m
Shot:
(c)
3.5  102 J
 1.7  102 N
2.00 m
Yes . If the muscles are capable of exerting 3.7  102 N on an object and giving that
object a kinetic energy of 7.4  102 J , as in the case of the shot, those same muscles
should be able to give the javelin a launch speed of
vi 
2 KE

m

  43 m
2 7.4  102 J
0.80 kg
s,
v 2  43 m s 
 i 
 1.9  102 m .
2
g 9.80 m s
2
with a resulting range of Rmax
Since this far exceeds the record range for the javelin, one must conclude that air
resistance plays a very significant role in these events.
165
C H A P T E R
5.73
5
The potential energy associated with the
wind is PEw  Fx , where x is measured
horizontally from directly below the pivot
of the swing and positive when moving
into the wind, negative when moving with
the wind. We choose PEg  0 at the level of
the pivot as shown in the figure. Also, note
that D  L sin   L sin 
D

so   sin 1   sin   , or
L

  sin 1 
50.0 m

 sin 50.0  28.94 .
 40.0 m

(a) Use conservation of mechanical energy, including the potential energy associated
with the wind. The final kinetic energy is zero if Jane barely makes it to the other

side, and KE  PEg  PEw
   KE  PE
 PEw
g
f

i
becomes
1
0  mg  L cos    F L sin    mvi2  mg  L cos    F L sin   ,
2
or v i 
2gL  cos   cos   
2FL
 sin   sin  
m
where m is the mass of Jane alone. This yields v i  6.15 m s .
(b) Again, using conservation of mechanical energy with KEf  0 ,
 KE  PE
g
 PEw
   KE  PE
f
g
 PEw

i
gives
1
0  Mg  L cos    F L sin    Mv i2  Mg  L cos    F L sin  
2
where M  130 kg is the combined mass of Tarzan and Jane. Thus,
vi 
2gL  cos   cos   
2FL
 sin   sin   which gives v i  9.87 m s
M
166
C H A P T E R
5.74
5
The power loss is the rate at which work is done moving air, or
loss
1
m v 2
W KE 2  
. But the mass of air moved in time t is



t
t
t
m   density  volume    area  length   A x  Av  t  .
Thus, loss 
 m  v 2   Av  t   v 2 1

  Av 3 .
2
2  t 
2  t 
When an object is moved at constant speed, the power expenditure is loss  fresistance v , so
the resistance force is given by
1
loss 1 2 Av

  Av 2 .
2
v
v
3
fresistance 
5.75
We choose PEg  0 at the level where the spring is relaxed (x = 0), or at the level of
position B.
(a) At position A, KE  0 and the total energy of the system is given by

E  0  PEg  PEs

1
 mg x1  k x12 , or
2
A


E   25.0 kg  9.80 m s 2  0.100 m  


1
2
2.50  104 N m  0.100 m   101 J
2
(b) In position C, KE  0 and the spring is uncompressed so PEs  0 .


Hence, E  0  PEg  0
or x 2 
C
 mg x 2
E
101 J

 0.410 m .
mg  25.0 kg  9.80 m s 2


1
(c) At Position B, PEg  PEs  0 and E   KE  0  0 B  mv B2 .
2
Therefore, v B 
2E

m
2 101 J
 2.84 m s .
25.0 kg
167
C H A P T E R
5
(d) Where the velocity (and hence the kinetic energy) is a maximum, the acceleration is
Fy
ay 
 0 (at this point, an upward force due to the spring exactly balances the
m
downward force of gravity). Thus, taking upward as positive, Fy  kx  mg  0 or
x 
mg
245 kg

 9.80  103 m  –9.80 mm .
k
2.50  104 N m
1
1
(e) From the total energy, E  KE  PEg  PEs  mv 2  mg x  k x 2 , we find
2
2
v
2E
k
 2 g x  x2
m
m
Where the speed, and hence kinetic energy is a maximum (i.e., at x   9.80 mm ),
this gives v max  2.85 m s
5.76
When the block moves distance x down the incline, the work done by the friction force
is W f   fk cos180 x   k nx   k  mg cos   x . From the work-kinetic energy theorem,

Wnc  KE  PEg  PEs
   KE  PE
f
g

 PEs , we find
i
Wnc  W f   k  mg cos   x  KE  PEg  PEs .
Since the block is at rest at both the start and the end, this gives
 k 19.6 N cos 37.0 0.200 m 
 0   19.6 N  0.200 m sin 37.0 
or
5.77
1
2
100 N m   0.200 m 

2
k  0.115 .
Choose PEg  0 at the level of the river. Then y i  36.0 m , y f  4.00 m , the jumper falls
32.0 m, and the cord stretches 7.00 m. Between the balloon and the level where the diver

stops momentarily, KE  PEg  PEs
   KE  PE
g
f
 PEs

i
gives
1
2
0  700 N  4.00 m   k 7.00 m   0  700 N  36.0 m   0 ,
2
or
k = 914 N m .
168
C H A P T E R
5.78
5
(a) Since the tension in the string is always perpendicular to the motion of the object,
the string does no work on the object. Then, mechanical energy is conserved:
 KE  PE    KE  PE  .
g
g
f
i
Choosing PEg  0 at the level where the string attaches to the cart, this gives
1
0  mg  L cos    mv o2  mg  L , or v 0 
2
2 g L 1  cos  
(b) If L = 1.20 m and   35.0 , the result of part (a) gives
v0 


2 9.80 m s 2 1.20 m 1  cos 35.0  2.06 m s .
169
C H A P T E R
5
Answers to Even Numbered Conceptual Questions
2.
(a) The chicken does positive work on the ground. (b) No work is done. (c) The crane does
positive work on the bucket. (d) The force of gravity does negative work on the bucket.
(e) The leg muscles do negative work on the individual.
4.
(a) Kinetic energy is always positive. Mass and speed squared are both positive.
(b) Gravitational potential energy can be negative when the object is below (i.e., closer to
the Earth) the chosen zero level.
6.
(a) Kinetic energy is proportional to the speed squared. Doubling the speed makes the
object’s kinetic energy four times larger. (b) If the total work done on an object in some
process is zero, its speed must be the same at the final point as it was at the initial point.
8.
The total energy of the bowling ball is conserved. Because the ball initially has
gravitational potential energy mgh and no kinetic energy, it will again have zero kinetic
energy when it returns to its original position. Air resistance and friction at the support
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.
10.
The effects are the same except for such features as having to overcome air resistance
outside. The person must lift his body slightly with each step on the tilted treadmill. Thus,
the effect is that of running uphill.
12.
Both the force of kinetic friction exerted on the sled by the snow and the resistance force
exerted on the moving sled by the air will do negative work on the sled. Since the sled is
maintaining constant velocity, some towing agent must do an equal amount of positive
work, so the net work done on the sled is zero.
14.
The kinetic energy is converted to internal energy within the brake pads of the car, the
roadway, and the tires.
16.
The kinetic energy is a maximum at the instant the ball is released. The gravitational
potential energy is a maximum at the top of the flight path.
18.
The normal force is always perpendicular to the surface and the motion is generally
parallel to the surface. Thus, in most circumstances, the normal force is perpendicular to
the displacement and does no work. The force of static friction does no work because
there is no displacement of the object relative to the surface in a static situation.
170
C H A P T E R
5
Answers to Even Numbered Problems
2.
30.6 m
4.
1.6 kJ
6.
0.68 J
8.
(a) 31.9 J
(b) 0
(c) 0 (d) 31.9 J
10.
160 m s
12.
(a) –168 J
14.
90.0 J
16.
(a) 1.2 J
18.
2.0 m
20.
(a) -19.6 J
22.
58.6 J
24.
26.5 m s
26.
5.1 m
28.
(a)
9.90 m s
30.
(a)
v B  5.94 m s , v C  7.67 m s
32.
5.11 m s
34.
61 m
36.
(a)
38.
1.2 kN
40.
77 m s
42.
(a)
2.29 m s
44.
h
4sin 2   1
5


(b) –184 J
(b)
(c) 500 J
(d) 148 J
(c) 6.3 J
5.0 m s
(b) 39.2 J
9.90 m s
(b)
(c) 0
7.67 m s
(b) 147 J
(b) 11.8 J
(b) 15.6 J
171
(e)
5.64 m s
C H A P T E R
46.
1.5 m (measured along the incline)
48.
(a) 21 kJ
50.
2.9 m s
52.
(a) 75.0 kJ
(b) 33.5 hp
54.
(a) 7.92 hp
(b) 14.9 hp
56.
(a) 7.50 J
(b) 15.0 J
58.
9.80 m s
60.
0.116 m
62.
1.4 m s
64.
3.9 kJ
66.
(a) 582 trips
(b) 90.5 W (0.121 hp)
68.
(a) 2.3 kN
(b)
4.7  102 N
70.
(a)
(b)
1.5  102 J
72.
(a)
76.
0.115
78.
(b)
3.1  102 J
5
(b) 0.92 hp
(c) 44.7 hp
(c) 7.50 J
(c) 0
(d) 30.0 J
(d)
1.5  102 J
3.5  102 J for javelin, 6.8  102 J for discus, 7.4  102 J for shot
(b) 1.7  102 N on javelin, 3.4  102 N on discus, 3.7  102 N on shot
(c) Yes
2.06 m s
172
173