Download Chapter 05 Solutions - Mosinee School District

Chapter 5
5
Energy
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
5.2
(a)
F cos
s
350 N cos 0
2.00 m
700 J
We assume the object moved upward with constant speed, so the kinetic energy did not change. Then, the
work–energy theorem gives the work done on the object by the lifter as
Wnc
KE
Wnc
(b)
PE
0
(mgy f
281.5 kg 9.80 m s2
mgyi )
mg
y , or
17.1 cm 1 m 102 cm
472 J
If the object moved upward at constant speed, the net force acting on it was zero. Therefore, the magnitude of
the upward force applied by the lifter must have been equal to the weight of the object
F
5.3
mg
281.5 kg 9.80 m s2
2.76
103 N
2.76 kN
Assuming the mass is lifted at constant velocity, the total upward force exerted by the two men equals the weight of
the mass: Ftotal
mg
653.2 kg 9.80 m s2
6.40
103 N . They exert this upward force through a total
upward displacement of 96 inches or 8 feet (4 inches per lift for each of 24 lifts). Thus, the total work done during
the upward movements of the 24 lifts is
W
5.4
(a)
F cos
x
Ftotal cos 0
x
6.40
10 5 N
cos 0
8 ft
1m
3.281 ft
2
10 4 J
The 35 N force applied by the shopper makes a 25° angle with the displacement of the cart (horizontal). The
work done on the cart by the shopper is then
Wshopper
F cos
x
35 N cos 25
50.0 m
Pages 5.1
1.6
103 J
Chapter 5
(b)
Since the speed of the cart is constant, KE f
KEi and Wnet
(c)
Since the cart continues to move at constant speed, the net work done on the cart in the second aisle is again
KE
0.
zero. With both the net work and the work done by friction unchanged, the work done by the shopper
Wshopper
Wnet
making F
F
5.5
(a)
Wfric is also unchanged. However, the shopper now pushes horizontally on the cart,
Wshopper ( x cos 0) = Wshopper
Wshopper
x smaller than before when the force was
x cos 35
The gravitational force acting on the object is
w
mg
5.00 kg 9.80 m s2
49.0 N
and the work done by this force is
Wg
PEg
mg y f
yi
w yi
yf
or
Wg
(b)
w L sin 30.0
49.0 N 2.50 m sin 30.0
The normal force exerted on the block by the incline is n
fk
kn
0.436 49.0 N cos 30.0
(c)
L
18.5 N cos180
2.50 m
= 180°), and the work it does is
46.3 J
Since the normal force is perpendicular to the displacement, the work done by the normal force is.
Wn
(d)
fk cos
mg cos 30.0 , so the friction force is
18.5 N
This force is directed opposite to the displacement (that is
Wf
61.3 J
n cos 90.0 L
0
If a shorter ramp is used to increase the angle of inclination while maintaining the same vertical displacement
Pages 5.2
Chapter 5
yf
yi , the work done by gravity will not change, the work done by the friction force will decrease
(because the normal force, and hence the friction force, will decrease and also because the ramp length L
decreases), and the work done by the normal force remains zero (because the normal force remains
perpendicular to the displacement).
5.6
(a)
WF
F
(b)
Since the crate moves at constant velocity, Thus, ax
Fx
x cos
150 N 6.00 m cos 0
0
fk
F
0
n
mg
fk
n
150 N
392 N
900 J
ay
0.
150 N
Also,
Fy
40.0 kg 9.80 m s2
392 N
so
k
5.7
(a)
Fy
F sin
n
mg
Fx
0.383
n
mg
0
F sin
F cos
kn
0
F cos
n
k
mg
F cos
F sin
k
k mg
F=
k
sin
cos
=
0.500 18.0 kg 9.80 m s2
0.500 sin 20.0
cos 20.0
Pages 5.3
= 79.4 N
Chapter 5
5.8
5.9
(b)
WF
(c)
fk
F cos
F cos
WF
F cos
WF
31.9 J
(b)
Wn
n cos 90 s
(c)
Wg
mg cos 90 s
(d)
Wnet
WF
(a)
The work–energy theorem, Wnet
(a)
W
s=
16.0 N cos 25.0°
(a)
(b)
KEi
1.49 kJ
Wn
F cos
10
1.49 kJ
2 .20 m
0
Wg
s
31.9 J
0
0
KE f
103 kg v2
F cos 0
3
103 J
1.49
0
Requiring that KEping pong
giving v
103 J
1.49
74.6 N cos180 º 20.0 m
s
1
2 .50
2
1
2 .45
2
20.0 m
74.6 N
fk cos
(b)
5.11
79.4 N cos 20.0º
Wf =
5 000 J
5.10
s
KEi , gives
0 , or
25.0 m
2.00 m s
v
5000 J , so F
KEbowling with KE
kg v2
31.9 J
1
2
200 N
mv2 , we have
1
7.00 kg 3.00 m s
2
2
160 m s .
1
2
mvi2 where vi2
giving
KEi
KE f
1
2
1
2
v02 x
v02 y
6.00 m s
5.75 kg 40.0 m2 s2
mv2f where v2f
v2x
v2y
2
2.00 m s
2
40.0 m2 s2 ,
115 J .
8.50 m s
Pages 5.4
2
5.00 m s
2
97.3 m2 s2 ,
Chapter 5
1
2
so , KE f
KE f
5.12
(a)
KEi
5.75 kg 97.3 m2 s2
280 J
115 J
280 J , and the change in kinetic energy has been
165 J
Since the applied force is horizontal, it is in the direction of the displacement giving
0 . The work done
by this force is then
WF0
F0 cos
x
F0 cos 0
x
F0
x
and
WF0
F0
(b)
350 J
12.0 m
x
29.2 N
Since the crate originally had zero acceleration, the original applied force was just enough to offset the
retarding friction force. Therefore, when the applied force is increased, it has a magnitude greater than the
friction force. This gives the crate a resultant force (and hence an acceleration) in the direction of motion,
meaning the speed of the crate will increase with time .
(c)
If the applied force is made smaller than F0, the magnitude of the friction force will be greater than that of the
applied force. This means the crate has a resultant force, and acceleration, in the direction of the friction force
(opposite to the direction of motion). The crate will now slow down and come to rest .
5.13
(a)
We use the work–energy theorem to find the work.
W
(b)
1
mv2f
2
KE
W
F cos
1
mvi2
2
0
s
fk cos180 s
W
k mg
5.6
1
70 kg 4.0 m s
2
kn
s
k
2
mg s ,
so
s
5.14
At the top of the arc, vy
102 J
0.70 70 kg 9.80 m s2
0 , and vx
v0 x
v0 cos 30.0
Pages 5.5
1.2 m
34.6 m s .
5.6
10 2 J
Chapter 5
Therefore, v2
1
mv2
2
KE
5.15
(a)
v2x
2
v2y
34.6 m s , and
1
0.150 kg 34.6 m s
2
2
90.0 J
As the bullet penetrates the tree trunk, the only force doing work on it is the force of resistance exerted by the
trunk. This force is directed opposite to the displacement, so the work done is
Wnet
KE f
f
(b)
( f cos180 ) x KE f KEi , and the magnitude of the force is
x cos180
(a)
KE A
1
mv2A
2
(b)
KEB
1
2
2.34
104 N
Wnet
2
x
vf
vi
2 . The time required to stop is then
2 5.50
0
10
2
m
575 m s
1
0.60 kg 2.0 m s
2
2 KE B
2 7.5 J
m
0.60 kg
KE
Wnet
Froad cos
Wnet
1 000 N
Also, Wnet
vi
2
1.91
10
4
s
1.2 J
mv2B , so
B
5.17
vf
x
v
t
(c)
2
If the friction force is constant, the bullet will have a constant acceleration and its average velocity while
stopping is v
5.16
1
7.80 10 3 kg 575 m s
2
5.50 10 2 m
0
KEi
KE f
KEB
1
s
KEA
Fresist cos
950 N 20 m
KEi
1
mv2
2
5.0 m s
7.5
2
s
1.0
1.2 J
6.3 J
1 000 N cos 0 s
103 J
0 , so
Pages 5.6
950 N cos180 s
Chapter 5
5.18
2 1.0
2 Wnet
m
v
1.0 m s
2000 kg
The initial kinetic energy of the sled is
1
mvi2
2
KEi
1
10 kg 2.0 m s
2
and the friction force is fk
Wnet
5.19
103 J
fk cos180 s
kn
KE f
k
2
20 J
mg
KEi , so s
9.8 N .
0 KEi
fk cos 180
20 J
9.8 N
2.0 m
With only a conservative force acting on the falling ball,
KE
PEg
KE
i
PEg
f
or
1
mvi2
2
Applying this to the motion of the ball gives 0
or
5.20
0.10 98 N
(a)
yi
1
mv2f
2
mgyi
mgyi
1
2
mv2f
mgyh
0
2
v2f
9.0 m s
2g
2 9.80 m s2
4.1 m
The force stretching the spring is the weight of the suspended object. Therefore, the force constant of the
spring is
Fg
k
(b)
mg
x
x
2.50 kg 9.80 m s2
2.76
10
2
m
8.88
10 2 N s
If a 1.25-kg block replaces the original 2.50-kg suspended object, the force applied to the spring (weight of
the suspended object) will be one-half the original stretching force. Since, for a spring obeying Hooke’s law,
the elongation is directly proportional to the stretching force, the amount the spring stretches now is
x
(c)
2
1
2
x
1
1
2.76 cm
2
1.38 cm
The work an external agent must do on the initially unstretched spring to produce an elongation xf is equal to
the potential energy stored in the spring at this elongation
Pages 5.7
Chapter 5
Wdone on
PEs
spring
5.21
PEs
f
1 2
kx f
2
i
1
8.88
2
0
102 N m 8.00
10
2
m
2
2.84 J
The magnitude of the force a spring must exert on the 3.65-g object to give that object an acceleration of
a
4.90 m s2 is given by Newton’s second law as
0.500g
F
ma
3.65
10
3
kg 4.90 m s2
1.79
10
2
N
Then, by Newton’s third law, this object exerts an oppositely directed force of equal magnitude in the on the spring.
If this reaction force is to stretch the spring 0.350 cm, the required force constant of the spring is
5.22
(a)
1.79 10 2 N
0.350 cm
F
x
k
10 2 N
10 -3 m
1.79
3.50
5.11 N m
While the athlete is in the air, the interacting objects are the athlete and Earth. They interact through the
gravitation force that one exerts on the other.
(b)
If the athlete leaves the trampoline (at the y = 0 level) with an initial speed of, vi
9.0 m s her initial
kinetic energy is
KEi
1
mvi2
2
1
60.0 kg 9.0 m s
2
2
and her gravitational potential energy is PEg
(c)
103 J
2.4
i
mgyi
mg 0
0J .
When the athlete is at maximum height, she is momentarily at rest and KE f
0 J Because the only force
acting on the athlete during her flight is the conservative gravitation force, her total energy (kinetic plus
potential) remains constant. Thus, the decrease in her kinetic energy as she goes from the launch point
where PEg
0 to maximum height is matched by an equal size increase in the gravitational potential
energy.
PEg
KE
PE f
PEi
KE f
and
PE f
0
2.4
103 J
0
2.4
103 J
Pages 5.8
KEi
or
PE f
PEi
KEi
KE f
Chapter 5
(d)
The statement that the athlete’s total energy is conserved is summarized by the equation
or KE2
1
2
PE2
mv22
1
2
mgy2
mv12
0
mgy1 . Solving for the final height gives
1
mv22
2
or y2
v12
y1
The given numeric values for this case are y1
v22
2g
0,
9.0 m s (at the trampoline level) and
1
0 at maximum height . The maximum height attained is then
v2
y2
(e)
PE
PE1 . In terms of mass, speed, and height, this becomes
1
mv2
2 1
mg
mgy1
y2
KE1
KE
v12
y1
v22
0
2g
9.0 m s
2
0
4.1 m
2 9.80 m s2
Solving the energy conservation equation given in part (d) for the final speed gives
2 1
mv2
m 2 1
v22
With y1
0 , v1
mgy1
mgy2
9.0 m s and y2
or
v12
v2
ymax 2
2g y1
y2
4.1 m 2 , the speed at half the maximum height is given
as
9.0 m s
v2
5.23
2
2 9.80 m s 2
0
4.1 m
2
6.4 m s
The work the beam does on the pile driver is given by
Wnc
F cos180
x
F 0.180 m
Here, the factor cos 180 is included because the force F exerted on the driver by the beam is directed upward, but
the
x
18.0 cm
0.180 m displacement undergone by the driver while in contact with the beam is directed
downward.
From the work–energy theorem, this work can also be expressed as
Pages 5.9
Chapter 5
Wnc
KE f
KEi
PE f
1
m v2f
2
PEi
vi2
mg y f
yi
Choosing y = 0 at the level where the pile driver first contacts the top of the beam, the driver starts from rest
0 at yi
vi
1
m 0
2
F 0.180 m
yielding
5.24
0 at y f
7.50 m and comes to rest again at v f
1.73 105 J
0.180 m
F
2 300 kg 9.80 m s2
0
9.62
0.180 m . Therefore, we have
0.180 m
7.50 m
105 N directed upward
When the child swings at the end of the ropes, she follows
a circular path of radius r = 1.75 as shown at the right. If we
choose y = 0 at the level of the child’s lowest position on this
path, then her y-coordinate when the rope is at
vertical is y
r
r cos
r 1
cos
angle with the
. Thus, her gravitational
potential energy at this position is
PEg
(a)
mgy
cos
wr 1
3.50
102 N 1.75 m 1
30.0 , PEg
3.50
(b)
When
(c)
At the bottom of the arc, y
cos
90.0 , and
When the ropes are horizontal,
PEg
5.25
mgr 1
cos 90.0
613 J
102 N 1.75 m 1
cos 30.0
0, cos = cos0° =1, and PEg
82.1 J .
0 .
While the motorcycle is in the air, only the conservative gravitational force acts on cycle and rider. Thus,
1
2
mv2f
y
mgy f
1
2
yf
yi
mvi2
mgyi , which gives
v2f
vi2
2g
35.0 m s
2
33.0 m s
2 9.80 m s2
2
6.94 m
Note that this answer gives the maximum height of the cycle above the end of the ramp, which is an unknown
distance above the ground.
Pages 5.10
Chapter 5
5.26
(a)
When equilibrium is reached, the total spring force supporting the load equals the weight of the load, or
Fs, total
Fs, leaf
Fs, helper
wload . Let k and kh represent the spring constants of the leaf spring and the
helper spring, respectively. Then, if x is the distance the leaf spring is compressed, the condition for
equilibrium becomes
kx
kh x
y0
wload
or
wload
k
x
(b)
105 N
5.00
kh y0
kh
5.25
3.60
105
Nm
10 5 N m
3.60
10 5
0.500 m
Nm
0.768 m
The work done compressing the springs equals the total elastic potential energy at equilibrium. Thus,
1
2
W
1
2
k x2
kh x
0.500 m
2
or
1
5.25
2
W
5.27
105 N m 0.768 m
2
1
2
3.60
10 5 N m 0.268 m
2
1.68
10 5 J
The total work done by the two bicep muscles as they contract is
Wbiceps
2Fav x
2 800 N 0.075 m
1.2
102 J
The total work done on the body as it is lifted 40 cm during a chin-up is
Wchin-up
Since Wchin-up
5.28
0.40 m
2.9
102 J
Wbiceps , it is clear that addition muscles must be involved .
Applying Wnc
Fm d
75 kg 9.80 m s2
mgh
KE
0
mgy f
PE
f
0
KE
PE
0
or
i
to the jump of the “original” flea gives
yf
Fm d
mg
where Fm is the force exerted by the muscle and d is the length of contraction.
If we scale the flea by a factor f, the muscle force increases by f 2 and the length of contraction increases by f. The
Pages 5.11
Chapter 5
mass, being proportional to the volume which increases by f 3 , will also increase by f 3 . Putting these factors into
our expression for y f gives
yf
f 2 Fm
super
flea
fd
f 3m
Fm d
mg
g
yf
0.5 m
so the “super flea” cannot jump any higher!
This analysis is used to argue that most animals should be able to jump to approximately the same height (~0.5 m).
Data on mammals from elephants to shrews tend to support this.
5.29
(a)
Taking y = 0, and hence PEg
mgy
0 at ground level, the initial total mechanical energy of the project
tile is
Etotal
KEi
i
PEi
1
mvi2
2
1
50.0 kg 1.20
2
(b)
10 2 m s
2
50.0 kg 9.80 m s 2 142 m
4.30
105 J
The work done on the projectile is equal to the change in its total mechanical energy.
Wnc
KE f
rise
PE f
KEi
1
50.0 kg 85.0 m s
2
3.97 10 4 J
(c)
mgyi
1
m v2f
2
PEi
2
120 m s
2
vi2
mg y f
yi
50.0 kg 9.80 m s 2
427 m
142 m
If, during the descent from the maximum height to the ground, air resistance does one and a half times as
much work on the projectile as it did while the projectile was rising to the top of the arc, the total work done
for the entire trip will be
Wnc
total
Wnc
2.50
Wnc
rise
3.97
descent
10 4 J
Wnc
9.93
rise
10 4
1.50 Wnc
rise
J
Then, applying the work–energy theorem to the entire flight of the projectile gives
Wnc
total
KE
PE
just before
hitting ground
KE
PE
at launch
and the speed of the projectile just before hitting the ground is
Pages 5.12
1
mv2f
2
mgy f
1
mvi2
2
mgyi
Chapter 5
vf
2 Wnc
total
vi2
m
2
2 g yi
10 4 J
9.93
(a)
2 9.80 m s2 142 m
0
115 m s
The work done by the gravitational force equals the decrease in the gravitational potential energy, or
Wg
(b)
2
120 m s
50.0 kg
5.30
yf
PE f
PEi
PEi
PE f
mg yi
yf
mgh
The change in kinetic energy is equal to the net work done on the projectile, which in the absence of air
resistance is just that done by the gravitational force. Thus,
(c)
5.31
Wnet
Wg
KE
KE
KE f
KEi
KE
mgh
so
mgh
KE f
KEi
mgh
1
mv02
2
mgh
(d)
No. None of the calculations in Parts (a), (b), and (c) involve the initial angle.
(a)
The system will consist of the mass, the spring, and Earth. The parts of this system interact via the spring
force, the gravitational force, and a normal force.
(b) The points of interest are where the mass is released from rest (at x = 6.00 cm) and the equilibrium point, x =
0.
(c)
The energy stored in the spring is the elastic potential energy, PEs
1
2
kx2 .
At x = 6.00 cm,
PEs
1
850 N m 6.00
2
10
2
m
2
1.53 J
and at the equilibrium position , (xf = 0)
PEs
(c)
1
k 0
2
2
0
The only force doing work on the mass is the conservative spring force (the normal force and the
gravitational force are both perpendicular to the motion). Thus, the total mechanical energy of the mass will
Pages 5.13
Chapter 5
be constant. Because we may choose y = 0, and hence PEg = 0, at the level of the horizontal surface, the
energy conservation equation becomes
KE f
PEs
f
KEi
PEs
i
1
mv2f
2
or
1 2
kx f
2
1
mvi2
2
1 2
kxi
2
and solving for the final speed gives
k 2
x
m i
vi2
vf
x 2f
If the final position is the equilibrium position x f
xi
0 at,
6.00 cm the final speed is
850 N m
1.00 kg
0
vf
(c)
0 and the object starts from rest vi
6.00
10
2
m
2
3.06 m 2 s 2
0
1.75 m s
When the object is halfway between the release point and the equilibrium position, we have
0, xi
vi
6.00 cm, and x f
850 N m
1.00 kg
0
vf
3.00 cm , giving
6.00
10
2
m
2
3.00
10
2
m
2
1.51 m s
This is not half of the speed at equilibrium because the equation for final speed is not a linear function of
position.
5.32
Using conservation of mechanical energy, we have
1
mv2f
2
mgy f
1
mvi2
2
0
or
yf
5.33
v2f
vi2
2g
10 m s
2
1.0 m s
2
5.1 m
2 9.80 m s2
Since no nonconservative 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
Pages 5.14
mv2f
mgy f
1
2
mvi2
mgyi with y f
0 yields
Chapter 5
v2f
yi
vi2
3.00 m s
2g
2
2 9.80 m
0
0.459 m
s2
Note that this result is independent of the mass of the child and sled.
5.34
(a)
The distance the spring is stretched is x
41.5 cm
0
35.0 cm
6.5 cm . Since the object hangs
in equilibrium on the end of the spring, the spring exerts an upward force of magnitude Fs = mg N on the
suspended object. Then, Hooke’s law gives the spring constant as
Fs
x
k
(b)
7.50 kg 9.80 m s 2
mg
6.5
0
2
10
1.1
m
103 N m
1.1 kN m
Consider one person (say the one holding the left end of the spring) to simply hold that end of the spring in
position while the person on the other end stretches it. The spring is then stretched a distance of
x
0.350 m by a stretching force of magnitude Fs = 190. From Hooke’s law, x
0
Fs k ,
we have
190 N
1.1 103 N m
0.350 m
5.35
(a)
0.17 m
and
0.17 m + 0.350 m
0.52 m
On a frictionless track, no external forces do work on the system consisting of the block and the spring as the
spring is being compressed. Thus, the total mechanical energy of the system is constant, or
KE f
PEg
f
PEs
f
KEi
PEg
i
PEs i . Because the track is horizontal, the gravitational
potential energy when the mass comes to rest is the same as just before it made contact with the spring, or
PEg
PEg . This gives
i
f
1
mv2f
2
(b)
1
mvi2
2
1 2
kxi
2
0 (the block comes to rest) and xi = 0 (the spring is initially undistorted),
Since v f
xf
1 2
kx f
2
vi
m
k
1.50 m s
0.250 kg
4.60 N m
0.350 m
If the track was not frictionless, some of the original kinetic energy would be spent overcoming friction
between the mass and track. This would mean that less energy would be stored as elastic potential energy in
Pages 5.15
Chapter 5
the spring when the mass came to rest. Therefore, the maximum compression of the spring would be less in
this case.
5.36
(a)
From conservation of mechanical energy,
1
mv2B
2
1
mv2A
2
mgyB
mgy A
or
v2A
vB
2g yA
yB
2 9.80 m s2 1.80 m
0
5.94 m s
Similarly,
v2A
vC
5.37
2g yA
2g 5.00 m
Wg
(a)
We choose the zero of potential energy at
C
PEg
0
(b)
A
PEg
yC
A
C
mg yA
yC
2.00 m
7.67 m s
49.0 N 3.00 m
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
mv2f
2
0
1
mvi2
2
mgyi
or
vf
(b)
vi2
2gyi
0
2 9.80 m s2
6.04 m
In this case, conservation of mechanical energy yields
Pages 5.16
10.9 m s
147 J
Chapter 5
vi2
vf
5.38
2gyi
2
At maximum height, vy = 0 and vx = v0 x
v2x
Thus, v f
v2y
energy gives PE f
(a)
2 9.80 m s2
40 m s cos 60
6.04 m
11.6 m s
20 m s .
20 m s . Choosing PEg = 0 at the level of the launch point, conservation of mechanical
KEi
v2f
vi2
yf
5.39
4.00 m s
KE f , and the maximum height reached is
40 m s
2g
2
20 m s
2
61 m
2 9.80 m s2
Initially, all the energy is stored as elastic potential energy within the spring. When the gun is fired, and as
the projectile leaves the gun, most of the energy in the form of kinetic energy along with a small amount of
gravitational potential energy. When the projectile comes to rest momentarily at its maximum height, all of
the energy is in the form of gravitational potential energy.
(b)
Use conservation of mechanical energy from when the projectile is at rest within the gun
vi
0, yi
0, and xi
yf
ymax
20.0 m , and x f
Then, KE
0
PEg
mgymax
0
0.120 m until it reaches maximum height where v f
PEs
f
0
0
0,
0 (the spring is relaxed after the gun is fired).
KE
PEg
PEs
i
becomes
1 2
kxi
2
or
2mgymax
xi2
k
(c)
2 20.0
10
3
kg 9.80 m s2
0.120 m
2
20.0 m
544 N m
This time, we use conservation of mechanical energy from when the projectile is at rest within the gun
vi
yf
0, yi
0, and xi
0.120 m, and x f
0.120 m until it reaches the equilibrium position of the spring
0 . This gives
Pages 5.17
Chapter 5
KE f
KE
PEg
k
xi2
m
v2f
PEs
Fy
or
f
1
mv2f
2
0
0
1 2
kx
2 i
mgy f
0
0.120 m
2
2 9.80 m s2
0.120 m
19.7 m s
yielding v f
(a)
PEs
2 gyf
544 N m
20.0 10 3 kg
5.40
PEg
i
0
n
F sin
mg
0 , or n
mg
F sin .
The friction force is then
fk
(b)
kn
k
mg
F sin
The work done by the applied force is
WF
F
x cos
Fx cos
and the work done by the friction force is Wfk
fk and x . Thus, Wfk
(c)
fk x cos180
fk
k
mg
x cos
where
F sin
is the angle between the direction of
x .
The forces that do no work are those perpendicular to the direction of the displacement
x.
These are n, mg, and the vertical coponent of F .
(d)
For part (a): n
fk
kn
mg
0.400 10.6 N
For part (b): WF
Wfk
5.41
(a)
F sin
fk x cos
Fx cos
2.00 kg 9.80 m s2
15.0 N sin 37.0
10.6 N
4.23 N
15.0 N 4.00 m cos 37.0
4.23 N 4.00 m cos180
47.9 J
16.9 J
When the child slides down a frictionless surface, the only nonconservative force acting on the child is the
normal force. At each instant, this force is perpendicular to the motion and, hence, does no work. Thus,
Pages 5.18
Chapter 5
conservation of mechanical energy can be used in this case.
(b)
The equation for conservation of mechanical energy, KE
1
2
m v2f
1
2
m gy f
m vi2
PE
f
KE
PE i , for this situation is
m gyi . Notice that the mass of the child cancels out of the equation, so the
mass of the child is not a factor in the frictionless case.
(c)
Observe that solving the energy conservation equation from above for the final speed gives
vi2
vf
2 g yi
yf
Since the child starts with the same initial speed ( vi
change in altitude in both cases,
(d)
f
0 ) and has the same
is the same in the two cases.
Work done by a non conservative force must be accounted for when friction is present. This is done by using
the work–energy theorem rather than conservation of mechanical energy.
(e)
From part (b), conservation of mechanical energy gives the final speed as
vi2
vf
5.42
(a)
2g yi
yf
2 9.80 m s2 12.0 m
0
15.3 m s
No. The change in the kinetic energy of the plane is equal to the net work done by all forces doing work on
it. In this case, there are two such forces, the thrust due to the engine and a resistive force due to the air. Since
the work done by the air resistance force is negative, the net work done (and hence the change in kinetic
energy) is less than the positive work done by the engine thrust. Also, because the thrust from the engine and
the air resistance force are nonconservative forces, mechanical energy is not conserved in this case.
(b)
Since the plane is in level flight, PEg
Wnc
Wthrust
F cos 0 s
Wresistance
KE f
f cos180 s
PEg
f
i
and the work–energy theorem reduces to
KEi , or
1
mv2f
2
1
mvi2
2
This gives
vf
vi2
2 F
f s
m
60 m s
2
2
Pages 5.19
7.5
4.0
104 N
1.5
104 kg
500 m
77 m s
Chapter 5
5.43
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 nonconservative force due to water resistance acts only during the last 5.0 m of fall. The work–energy theorem
then gives
Wnc
KE
PEg
KE
f
PEg
i
or
Fav cos180
5.0 m
0
0
70 kg 9.80 m s2 15 m
0
This gives the average resistance force as Fav
5.44
(a)
103 N
2.1
2.1 kN .
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
vf
(b)
PEg
KE
f
PEg , or
i
2 9.80 m s2
2gyi
1
2
mv2f
0
0.268 m
0
mgyi , which gives
2.29 m s
With a nonconservative force present, we use
Wnc
KE
PEg
f
KE
PEg
i
1
mv2f
2
0
0
mgyi
or
Wnc
m
v2f
2
25.0 kg
gyi
2.00 m s
2
2
9.80 m s2
Thus, 15.7 J of energy is spent overcoming friction.
Pages 5.20
0.268 m
15.7 J
Chapter 5
5.45
Choose PEg = 0 at the level of the bottom of the driveway.
Then Wnc
KE
PEg
KE
f
1
mv2f
2
f cos 180° s
0
PEg
0
i
becomes
mg s sin 20°
Solving for the final speed gives
2f s
m
vf
2 gs sin 20
vf
2 9.80 m s2
or
5.46
(a)
5.0 m sin 20
2 4.0
103 N 5.0 m
2.10
103 kg
3.8 m s
Yes. Two forces, a conservative gravitational force and a nonconservative normal force, act on the child as
she goes down the slide. However, the normal force is perpendicular to the child’s motion at each point on the
path and does no work. In the absence of work done by nonconservative forces, mechanical energy is
conserved.
(b)
We choose the level of the pool to be the y = 0 (and hence, PEg = 0) level. Then, when the child is at rest at
the top of the slide, PEg
the child as. Etotal
(c)
PEg
mgy
PEg
0 and KE
KE
mgh and KE
PEg
mgh 5 and KE
0 . Note that this gives the constant total mechanical energy of
mgh At the launching point (where y = h/5), we have
Etotal
PEg
4mgh 5 . At the pool level,
mgh .
At the launching point (i.e., where the child leaves the end of the slide),
KE
1
mv02
2
4mgh
5
meaning that
v0
(d)
8 gh
5
After the child leaves the slide and becomes a projectile, energy conservation gives
Pages 5.21
Chapter 5
KE
1
2
PEg
mv2
mgy
mgh where v2
Etotal
varies with time. At maximum height, y
1
m v02 x
2
(e)
0
mgymax
ymax and vy
ymax
and
mgh
v2x
h
v2y . Here, vx
v0 x is constant, but v y
0 , yielding
v02 x
2g
v0 cos . Substituting this into the result from
If the child’s launch angle leaving the slide is , then v0 x
part (d) and making use of the result from part (c) gives
ymax
(f)
v02
cos2
2g
h
h
1 8 gh
cos2
2g
5
or
ymax
h 1
4
cos2
5
No. If friction is present, mechanical energy would not be conserved, so her kinetic energy at all points after
leaving the top of the waterslide would be reduced when compared with the frictionless case. Consequently,
her launch speed, maximum height reached, and final speed would be reduced as well.
5.47
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, n2
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
k
mg cos 10.5° 200 m
Applying Wnc
k
cos 180°
KE
PEg
200 m
f
mg cos10.5° 200 m
fk
cos 180° x
2
mg x
k
KE
k
PEg
i
to this complete trip gives
mg x
0
0
0
mg 200 m sin 10.5°
or
x
sin 10.5
k
cos 10.5°
200 m . If
k
0.0750 , then x
Pages 5.22
289 m .
Chapter 5
5.48
The normal force exerted on the sled by the track is n
fk
kn
k
mg cos
and the friction force is
mg cos .
If s is the distance measured along the incline that the sled travels, applying
Wnc
KE
k
PEg
mg cos
f
KE
PEg
cos 180° s
i
0
to the entire trip gives
1
mvi2
2
mg s sin
0
or
s
5.49
4.0 m s
vi2
2 g sin
k
2 9.80 m s2
cos
(a) Consider the entire trip and apply Wnc
f1 cos 180° d1
f2 cos 180° d2
2
sin 20
KE
1
mv2f
2
PEg
0
1.5 m
0.20 cos 20
KE
f
0
PEg
i
to obtain
mgyi
or
vf
f1d1
2
g yi
2
9.80 m s2
which yields v f
f 2 d2
m
3 600 N 200 m
80.0 kg
24.5 m s .
(b)
Yes, this is too fast for safety.
(c)
Again, apply Wnc
d1
1 000 m
KE
d2 , and v f
f1 cos 180° 1 000 m
which reduces to
50.0 N 800 m
1 000 m
PEg
KE
f
PEg , now with d2 considered to be a variable,
i
5.00 m s . This gives
d2
1 000 m f1
1
mv2f
2
f2 cos 180° d2
f1 d2
f2 d2
Pages 5.23
1
2
mv2f
0
0
mgyi
mgyi . Therefore,
Chapter 5
mg yi
d2
1
mv2f
2
1 000 m f1
f2
f1
784 N 1 000 m
1
80.0 kg 5.00 m s
2
1 000 m 50.0 N
3 600 N
2
206 m
50.0 N
(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.
5.50
Wnc
(a)
KE
y
60 m sin30
0 because the speed is constant. The skier rises a vertical distance of
30 m . Thus,
30 m
104 J
2.06
21 kJ
The time to travel 60 m at a constant speed of 2.0 m/s is 30 s. Thus, the required power input is
Wnc
t
P
5.51
KE
70 kg 9.80 m s2
Wnc
(b)
PE , but
2.06 10 4 J
30 s
686 W
1 hp
746 W
0.92 hp
As the piano is lifted at constant speed up to the apartment, the total work that must be done on it is
Wnc
KE
PEg
0
mg y f
yi
3.50
103 N 25.0 m
8.75
104 J
The three workmen (using a pulley system with an efficiency of 0.750) do work on the piano at a rate of
Pnet
0.750 3Psingle
0.750 3 165 W
371 W
371 J s
worker
so the time required to do the necessary work on the piano is
t
5.52
Wnc
Pnet
8.75 10 4 J
371 J s
236 s
236 s
1 min
60 s
3.93 min
Let N be the number of steps taken in time t. We determine the number of steps per unit time by
Pages 5.24
Chapter 5
Power =
work per step per unit mass mass # steps
work done
t
t
or
70 W
0.60
J step
kg
N
t
60 kg
, giving
N
= 1.9 steps s
t
The running speed is then
av
5.53
x
t
N
t
distance traveled per step
Assuming a level track, PE f
Wnc
KE
PE
1
m v2f
2
vi2
step
s
1.5
m
step
2.9 m s
PEi , and the work done on the train is
KE
f
1.9
PE
i
1
0.875 kg
2
0.620 m s
2
0
0.168 J
The power delivered by the motor is then
P
5.54
Wnc
t
0.168 J
21.0 10 3 s
8.01 W
When the car moves at constant speed on a level roadway, the power used to overcome the total frictional force
equals the power input from the engine, or
ftotal
5.55
Pinput
v
175 hp 746 W
29 m s 1 hp
Poutput
4.5
ftotalv
105 N
The work done on the older car is
Wnet
old
KE f
KEi
old
1
mv2
2
0
1
mv2
2
The work done on the newer car is
Pages 5.25
Pinput . This gives
Chapter 5
Wnet
KE f
new
KEi
1
m 2v
2
new
2
0
4
1
mv2
2
4 Wnet
old
and the power input to this car is
Wnet
Pnew
4 Wnet
new
t
old
4Pold
t
or the power of the newer car is 4 times that of the older car .
5.56
Neglecting any variation of gravity with altitude, the work required to lift a 3.20
y
to an altitude of
W
PEg
1.75 km is
mg
y
3.20
107 kg 9.80 m s2 1.75
The time required to do this work using a
5.57
(a)
5.49 1011 J
2.70 103 J s
W
t
P
107 kg load at constant speed
2.03
P
2.70 kW
108 s
2.03
103 m
5.49
1011 J
103 J s pump is
2.70
108 s
1h
3 600 s
5.64
10 4 h
The acceleration of the car is
a
v
v0
t
18.0 m s 0
12 .0 s
1.50 m s2
Thus, the constant
forward force due to the engine is found from
Fengine
Fair
ma
400 N
1.50
F
Fengine
103 kg 1.50 m s2
The average velocity of the car during this interval is v
av
input from the engine during this time is
Pages 5.26
ma as
Fair
v
2 .65
v0 / 2
103 N
9.00 m s , so the average power
Chapter 5
Pav
(b)
Fengine vav
2.65
103 N 9.00 m s
1 hp
746 W
At t = 12.0 s, the instantaneous velocity of the car is v
32.0 hp
18.0 m s and the instantaneous power input from
the engine is
P
5.58
(a)
Fengine v
2.65
1 hp
746 W
103 N 18.0 m s
63.9 hp
The acceleration of the elevator during the first 3.00 s is
v
a
1.75 m s
3.00 s
v0
t
so F
net Fmotor
Fmotor
0.583 m s2
ma gives the force exerted by the motor as
mg
m a
0
g
650 kg
0.583
9.80
m s2
6.75
103 N
(v v0 ) 2
The average velocity of the elevator during this interval is vav
0.875 m s so the average
power input from the motor during this time is
Pav
Fmotor vav
6.75
103 N 0.875 m s
1 hp
746 W
(b) When the elevator moves upward with a constant speed of v
motor is F
motor
P
5.59
mg v
7.92 hp
1.75 m s , the upward force exerted by the
mg and the instantaneous power input from the motor is
650 kg 9.80 m s2 1.75 m s
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 .
Pages 5.27
1 hp
746 W
14.9 hp
Chapter 5
(a)
For x
0 to x
W
1
8.00 m 6.00 N
2
8
For to x
W8
5.60
W0
altitude
24.0 J
10.0 m ,
10
1
CE
2
area of triangle CDE
10
1
2.00 m
2
=
(c)
1
AC
2
area of triangle ABC
W0
(b)
8.00 m ,
W0
8
W8
3.00 N
3.00 J
24.0 J
10
altitude
3.00 J
21.0 J
The work done by a force equals the area under
the force versus displacement curve.
(a)
For the region 0
5.00 m ,
3.00 N 5.00 m
W0 to 5
(b)
x
2
For the region 5.00 m
W5 to 10
x xf
1
mv2f
2
10.0 m ,
3.00 N 5.00 m
(c) For the region 10.0 m
(d) KE
x
KE
x 0
1
mv02
2
x
7.50 J
15.0 J
15.0 m , W10 to 15
W0 to x f
3.00 N 5.00 m
area under F vs. x curve from x
W0 to x f
giving
vf
v02
2
2
W0 to x f
m
Pages 5.28
7.50 J
0 m to x
xf
, or
Chapter 5
For x f
5.00 m :
For x f
2
W0 to 5
m
v02
vf
0.500 m s
2
3.00 kg
2
7.50 J
2.29 m s
15.0 m :
2
W0 to 15
m
v02
vf
2
m
v02
W0 to 5
W5 to 10
W10 to 15
or
0.500 m s
vf
5.61
(a)
(b)
Fx
8x
2
2
3.00 kg
7.50 J
15.0 J
7.50 J
4.50 m s
16 N See the graph at the right.
The net work done is the total area under the graph
from x
0 to x
3.00 m . This consists of two
triangular shapes, one below the axis (negative area)
and one above the axis (positive). The net work is then
2 .00 m
Wnet
16.0 N
1.00 m 8.00 N
2
2
12.0 J
5.62
(b)
Solving part (b) first, we recognize that the egg has constant acceleration ay
g as it falls 32.0 m from
rest before contacting the pad. Taking upward as positive, its velocity just before contacting the pad is given
by v2
v1
v02
2a y
0
2
y as
9.80 m s2
32.0 m
25.0 m s
The average acceleration as the egg comes to rest in 9.20 ms after contacting the pad is
Pages 5.29
Chapter 5
vf
aav
0
v1
t
25.0 m s
9.20
3
10
25.0
9.20
s
103 m s2
and the average net force acting on the egg during this time is
Fnet
(a)
maav
av
75.0
10
3
kg
25.0
9.20
103 m s2
204 N
204 N upward
y as the upward force Fnet
The egg has a downward displacement of magnitude
av
brings the egg to rest.
The net work done on the egg in this process is
Wnet
Fnet
av
cos180°
y
KE f
75.0
10
KE1
0
1
mv12
2
so
y
5.63
mv12
cos180
av
2 Fnet
3
kg 25.0 m s
2 204 N
2
0.115 m
11.5 cm
The person’s mass is
m
w
g
700 N
9.80 m s2
71.4 kg
The net upward force acting on the body is F
net
2 355 N
700 N
10.0 N . The final upward velocity can
then be calculated from the work–energy theorem as
Wnet
KE f
KEi
1
mv2
2
1
mvi2
2
or
Fnet cos
which gives v
s
10.0 N cos 0°
0.250 m
0.265 m s upward .
Pages 5.30
1
71.4 kg v2
2
0
Chapter 5
5.64
Taking y = 0 at ground level, and using conservation of energy from when the boy starts from rest (v
i
top of the slide ( y
i
velocity (v0 x
H ) to the instant he leaves the lower end ( y f
v f , v0 y
1
mv2f
2
mgh
0) at the
h) of the frictionless slide with a horizontal
0) , yields
0
v2f
or
mgH
2g H
h
[1]
Considering his flight as a projectile after leaving the end of the slide,
y
v0 y t
1
2
ay t 2 gives the time to drop
distance h to the ground as
h
1
2
0
g t2
or
2h
g
t
The horizontal distance traveled (at constant horizontal velocity) during this time is d, so
d
v0x t
vf
2h
g
and
d
vf
gd 2
2h
g
2h
Substituting this result into Equation [1] above gives
gd 2
2h
5.65
(a)
(b)
2g H
h
If y = 0 at point B, then yA
PEA
mgyA
1.50
PEB
mgyB
0 and
If y = 0 at point C, then yA
or
H
h
d2
4h
35.0 m sin 50.0°
103 kg 9.80 m s2
PEA
B
PEB
26.8 m and y
B
26.8 m
PEA
50.0 m sin 50.0°
0
3.94
3.94
38.3 m and yB
0 . Thus,
105 J
105 J
mgyA
1.50
103 kg 9.80 m s2
Pages 5.31
38.3 m
5.63
105 J
15.0 m sin 50.0°
this case,
PEA
3.94
105 J
11.5 m . In
Chapter 5
PEB
mgyB
103 kg 9.80 m s2 11.5 m
1.50
1.69
10 5 J
and
PEA
5.66
PEB
B
PEA
105 J
1.69
5.63
105 J
3.94
105 J
The support string always lies along a radius
line of the circular path followed by the bob.
This means that the tension force in the string
is always perpendicular to the motion of the
bob and does no work. Thus, mechanical energy
is conserved and (taking y = 0 at the point of
support) this gives
1
2
1
2
mv2
mvi2
mg yi
yf
0
mg
L cos
L
i
or
5.67
2gL 1
v
1.9 m s
(a)
The equivalent spring constant of the bow is given F
k
(b)
cos
2 9.8 m s2
v
Ff
i
2.0 m 1
cos 25°
kx by as
230 N
= 575 N m
0.400 m
xf
From the work–energy theorem applied to this situation,
Wnc
KE
PEg
PEs
f
KE
PEg
PEs
i
The work done pulling the bow is then
Wnc
1 2
kx f
2
1
575 N m 0.400 m
2
2
46.0 J
Pages 5.32
0
0
1 2
kx
2 f
0
0
0
Chapter 5
5.68
Choose PEg = 0 at the level where the block comes to rest against the spring. Then, in the absence of work done by
nonconservative forces, the conservation of mechanical energy gives
KE
PEg
PEs
KE
f
PEg
PEs
i
or
0
1 2
kx f
2
0
0
mg L sin
0
Thus,
5.69
(a)
From v2
v
(b)
2 12.0 kg 9.80 m s2
2 mg L sin
k
xf
v02
3.00
3.00 m sin 35.0
0.116 m
Nm
y , we find the speed just before touching the ground as
2a y
2 9.80 m s2 1.0 m
0
104
4.4 m s
Choose PEg = 0 at the level where the feet come to rest. Then
Wnc
KE
PEg
Fav cos 180° s
KE
f
0
PEg
becomes
i
1
mvi2
2
0
mg s
or
Fav
5.70
mvi2
2s
mg
75 kg 4.4 m s
2 5.0
10
3
2
75 kg 9.80 m s2
m
From the work–energy theorem,
Wnc
KE
PEg
PEs
f
KE
PEg
PEs
Pages 5.33
i
1.5
105 N
Chapter 5
we have
1
mv2f
2
fk cos 180° s
0
0
0
1 2
kx
2 i
0
or
f
5.71
(a)
kxi2
2 fk s
m
10 2 m)2
5.3 10
(8.0 N/m)(5.0
The two masses will pass when both are at y f
of energy, KE
1
m1
2
PEg
m2 v2f
PEs
m1
KE
f
m2 gy f
2(0.032 N)(0.15 m)
kg
1.4 m/s
2.00 m above the floor. From conservation
PEg
0
3
0
PEs
m1gy1i
i
0
or
vf
2 m1 g y1i
m1 m2
2 gyf
2 5.00 kg 9.80 m s2
4.00 m
2 9.80 m s2
8.00 kg
3.13 m s .
This yields the passing speed as v f
(b)
When m
1
Thus, KE
1
m1
2
5.00 kg reaches the floor, is above the floor, m2
PEg
m2 v2f
PEs
f
m2 g y2 f
KE
0
2 .00 m
PEg
0
PEs
m1gy1i
or
Pages 5.34
i
0
becomes
3.00 kg
IS
y2 f
4.00 m above the floor.
Chapter 5
2 g m1 y1i
vf
m2 y2 f
m1
m2
Thus,
2 9.80 m s2
vf
(c)
3.00 kg 4.00 m
When the 5.00-kg hits the floor, the string goes slack and the 3.00-kg becomes a projectile launched straight
y
v2y
v02 y
2 ay
4.43 m s . At the top of its arc, vy
0
4.43 m s
2
9.80 m s2
k nA
yB
vertical displacement of block B is
v02 y
2ay
y gives
1.00 m
mA g cos 37 , and the friction force is
. The vertical distance block A rises is
k mA g cos 37
0 and v2y
2
The normal force the incline exerts on block A is nA
fk
4.43 m s
8.00 kg
upward with initial speed v0 y
5.72
5.00 kg 4.00 m
yA
20 m sin 37°
yA
mB g
12 m , while the
20 m .
We find the common final speed of the two blocks by use of
Wnc
KE
k mA g
This gives
v2f
PEg
2g
f
cos 37° s
mB
yB
KE
PEg
1
m
2 A
mA
yA
mA
2 9.80 m s2
100 kg
KE
i
mB v2f
k mA
PEg
0
yB
, or
cos 37° s
mB
20 m
50 kg 12 m
150 kg
which yields v2f
mA g
157 m2 s2 .
Pages 5.35
0.25 50 kg 20 m cos 37°
Chapter 5
The change in the kinetic energy of block A is then
KE A
5.73
1
mAv2f
2
1
50 kg 157 m 2 s2
2
0
103 J
3.9
3.9 kJ
Since the bowl is smooth (that is, frictionless),
mechanical energy is conserved or
KE
PE
KE
f
PE
i
Also, if we choose y = 0 (and hence, PEg = 0)
at the
lowest point in the bowl, then
yA
(a)
5.74
R, yB
PEg
A
PEg
A
0, and yC
mgyA
mgR , or
0.200 kg 9.80 m s2
(b)
KEB
KEA
(c)
KE B
1
2
(d)
PEg
C
2R 3 .
PEA
mv2B
PEB
vB
0
0.300 m
mgyA
mgyB
PEB
PEC
0.588 J
2 0.588 J
2 KE B
m
0.200 kg
0.200 kg 9.80 m s 2
mgyC
0.588 J
KEC
KEB
0.588 J
0
(a)
KEB
1
mv2B
2
(b)
The change in the altitude of the particle as
1
0.200 kg 1.50 m s
2
2
2 0.300 m
3
0.392 J
0.225 J
Pages 5.36
0
0.588 J
2.42 m s
0.392 J
0.196 J
Chapter 5
it goes from A to B is yA
yB
R , where
R = 0.300 m is the radius of the bowl. Therefore,
the work–energy theorem gives
Wnc
KE B
KE B
KE A
PE B PE A
0 mg yB y A
KE B
mg
R
or
Wnc
0.200 kg 9.80 m s2
0.225 J
0.300 m
0.363 J
The loss of mechanical energy as a result of friction is then .0.363 J.
(c)
No. Because the normal force, and hence the friction force, vary with the position of the particle on its path,
it is not possible to use the result from part (b) to determine the coefficient of friction without using calculus.
5.75
(a)
Consider the sketch at the right. When the mass m = 1.50 kg is in
equilibrium, the upward spring force exerted on it by the lower spring
(i.e., the tension in this spring) must equal the weight of the object, or
FS 2
x2
mg . Hooke’s law then gives the elongation of this spring as
FS 2
k2
mg
k2
Now, consider point A where the two springs join. Because this point is
in equilibrium, the upward spring force exerted on A by the upper spring
must have the same magnitude as the downward spring force exerted on A
by the lower spring (that is, the tensions in the two springs must be equal).
The elongation of the upper spring must be
x1
FS1
k1
FS 2
k1
mg
k1
and the total elongation of the spring system is
Pages 5.37
Chapter 5
x
x1
mg
k1
x2
mg
k2
1.50 kg 9.80 m s2
(b)
mg
1
k1
1
k2
1
103 N m
1.20
1.80
The spring system exerts an upward spring force of FS
1
103 N m
2.04
10
2
m
mg on the suspended object and undergoes an
elongation of x. The effective spring constant is then
keffective
5.76
FS
x
mg
x
1.50 kg 9.80 m s2
2.04
10
2
7.20
m
102 N m
Refer to the sketch given in the solution of Problem 5.75.
(a)
Because the object of mass m is in equilibrium, the tension in the lower spring,
FS 2
must equal the weight of
the object. Therefore, from Hooke’s law, the elongation of the lower spring is
x2
FS 2
k2
mg
k2
From the fact that the point where the springs join (A) is in equilibrium, we conclude that the tensions in the
two springs must be equal, FS1
FS2
x1
mg . The elongation of the upper spring is then
FS1
k1
mg
k1
and the total elongation of the spring system is
x
(b)
x1
x2
mg
1
k1
1
k2
The two spring system undergoes a total elongation of x and exerts an upward spring force FS
suspended mass. The effective spring constant of the two springs in series is then
Pages 5.38
mg on the
Chapter 5
keffective
5.77
(a)
FS
x
mg
x
mg
mg
1
k1
1
k1
1
k2
1
k2
1
The person walking uses Ew = (220 kcal)(4 186 J/1 kcal = 9.21
105 J of energy while going 3.00 miles. The
quantity of gasoline which could furnish this much energy is
V1
9.21 105 J
1.30 108 J gal
7.08
10
3
gal
This means that the walker’s fuel economy in equivalent miles per gallon is
fuel economy
(b)
3.00 mi
7.08 10 3 gal
423 mi gal
In 1 hour, the bicyclist travels 10.0 miles and uses
EB
400 kcal
4 186 J
1 kcal
106 J
1.67
which is equal to the energy available in
V2
1.67 106 J
1.30 108 J gal
1.29
10
2
gal
of gasoline. Thus, the equivalent fuel economy for the bicyclist is
10.0 mi
1.29 10 2 gal
5.78
776 mi gal
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
Pages 5.39
Chapter 5
Each time the student climbs the stairs, she raises her body a vertical distance of
y
80 steps 0.150 m step
50.0 kg 9.80 m s2
PEg
(a)
12.0 m . The mechanical energy required to do this is
12.0 m
1 kcal
4186 J
103 J
5.88
PEg
mg
y , or
1.40 kcal
The number of times the student must climb the stairs to metabolize 1 pound of fat is
Em
PEg
N
817 kcal
1.40 kcal trip
582 trips
It would be more practical for her to reduce food intake.
(c)
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
Pav
5.79
(a)
5.88 103 J
65.0 s
W
t
90.5 W
Use conservation of mechanical energy, (KE
90.5 W
1 hp
746 W
PEg ) f
(KE
to find the speed of the skier as he leaves the track. This gives
2g( yi
yf
2(9.80 m/s 2 )(4.00 m)
1
2
0.121 hp
PEg )i , from the start to the end of the track
mv2
mgy f
28.0 m s cos 45.0°
19.8 m s .Thus , vtop
v2x
v2y
mechanical energy from the end of the track to the top of the arc gives
1
2
m 19.8 m s
ymax
2
10.0 m
mg ymax
1
2
m 28.0 m s
28.0 m s
2
2
19.8 m s
mg 10.0 m , or
2
2 9.80 m s2
Pages 5.40
mgyi , or
28.0 m/s
(b) At the top of the parabolic arc the skier follows after leaving the track, and vy
vx
0
30.0 m
0 . Thus,
19.8 m s Applying conservation of
Chapter 5
(c)
y
Using
1
2
v0 y t
10.0 m
ay t 2 for the flight from the end of the track to the ground gives
1
2
28.0 m s sin 45.0° t
9.80 m s2 t 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
5.80
v0 x t
28.0 m s cos 45.0°
4.49 s
89.0 m
First, determine the magnitude of the applied force by considering
a free-body diagram of the block. Since the block moves with
constant velocity,
From
Fx
Thus, fk
Fx
0 , we see that n
kn
F sin 30
0.
Fy
F cos 30 .
k F cos 30
mg
, and
Fy
0 becomes
k F cos 30
or
F
(a)
sin 30°
k
cos 30°
sin 30°
0.30 cos 30°
2 .0
10 2 N
The applied force makes a 60° angle with the displacement up the wall. Therefore,
WF
5.81
5.0 kg 9.80 m s2
mg
F cos 60° s
2.0
102 N cos 60°
49 N
1.0 3.0 m
(b)
Wg
mg cos180° s
(c)
Wn
n cos 90 s
(d)
PEg
mg
y
3.0 m
3.1
1.5
102 J
102 J
0
49 N 3.0 m
1.5
102 J
We choose PEg = 0 at the level where the spring is relaxed (x = 0), or at the level of position B.
Pages 5.41
Chapter 5
(a)
(b)
At position A, KE = 0 and the total energy of the system is given by
E
0
PEg
PEs
mgx1
E
25.0 kg 9.80 m s2
A
1
k x12 , or
2
1
2.50
2
0.100 m
10 4 N m
0.100 m
2
101 J
In position C, and the spring is uncompressed, so PEs = 0. Hence,
E
0
PEg
0
mg x2
C
or
E
mg
x2
(c)
101 J
0.410 m
25.0 kg 9.80 m s2
At position B, PEg
0 and E
PEs
KE
0
0
B
1
2
mv2B
Therefore,
vB
(b)
2 101 J
2E
m
25.0 kg
2.84 m s
Where the velocity (and hence the kinetic energy) is a maximum, the acceleration is ay
( Fy ) m
0 (at
this point, an upward force due to the spring exactly balances the downward force of gravity). Thus, taking
Fy
upward as Positive,
x
(e)
mg
k
2.50
kx
245 kg
10 4 N m
From the total energy, E
v
2E
m
2 gx
mg
KE
PEg
0
9.80
10
PEs
1
2
3
m
mv2
9.80 mm
mgx
1
2
kx2 , we find
k 2
x
m
Where the speed, and hence kinetic energy is a maximum (that is, at x = 9.80 mm), this gives
Pages 5.42
Chapter 5
2 101 J
vmax
25.0 kg
2 9.80 m s2
9.80
10
3
104 N m
2.50
m
9.80
25.0 kg
10
3
m
2
or
vmax
5.82
2.85 m s
When the hummingbird is hovering, the magnitude of the average upward force exerted by the air on the wings
(and hence the average downward force the wings exert on the air) must be Fav
mg where mg is the weight of
the bird. Thus, if the wings move downward distance d during a wing stroke, the work done each beat of the wings
is
Wbeat
Fav d
mgd
3.0
3
10
kg 9.80 m s2
3.5
10
2
m
1.0
10
3
J
In 1 minute, the number of beats of the wings that occur is
N
80 beats s 60 s min
103 beats min
4.8
so the total work preformed in 1 minute is
Wtotal
5.83
Choose PEg
NWbeat 1 min
4.8
103
beats
min
0 at the level of the river. Then yi
1.0
10
36.0 m , y f
3
J
beat
1 min
4.9 J
4.00 , 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
f
KE
0
700 N 4.00 m
k
914 N m
PEg
PEs gives
i
1
k 7.00 m
2
2
0
700 N 36.0 m
0
or
5.84
If a projectile is launched, in the absence of air resistance, with speed v 0 at angle
required to return to the original level is found from
y
Pages 5.43
v0 y t
1
2
ay t 2 as 0
above the horizontal, the time
v0 sin
t
(g 2) t 2 , which
Chapter 5
gives t
(2 v0 sin ) g . The range is the horizontal displacement occurring in this time.
Thus,
R
v0 x t
v0 cos
2 v0 sin
g
Maximum range occurs when
v02 2 sin cos
v02 sin 2
g
= 45 , giving or Rmax
g
v02 g .or v02
g Rmax The minimum kinetic energy
required to reach a given maximum range is then
KE
(a)
1
mv02
2
The minimum kinetic energy needed in the record throw of each object is
Javelin:
KE
Discus: KE
Shot:
(b)
1
mg Rmax
2
KE
1
0.80 kg 9.80 m s2
2
1
2.0 kg 9.80 m s2
2
1
7.2 kg 9.80 m s2
2
98 m
74 m
23 m
3.8
7.3
8.1
10 2 J
10 2 J
10 2 J
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 as Wnet
Fav s
each object is
Javelin:
Fav
3.8 102 J
2.00 m
1.9
10 2 N
Discus:
Fav
7.3 102 J
2.00 m
3.6
102 N
Pages 5.44
KE as Fav
KE
0 s Thus, the required force for
Chapter 5
(c)
8.1 10 2 J
2.00 m
Fav
Shot:
10 2 N
4.1
Yes. If the muscles are capable of exerting 4.1
102 N on an object and giving that object a kinetic energy of
102 J, as in the case of the shot, those same muscles should be able to give the javelin a launch speed of
8.1
2 8.1
2 KE
m
v0
102 J
45 m s
0.80 kg
with a resulting range of
Rmax
45 m s
v02
g
2
9.80 m s2
2.1
102 m
Since this far exceeds the record range for the javelin, one must conclude that air resistance plays a very
significant role in these events.
5.85
From the work–energy theorem, Wnet
KE f
KEi .giving Wnet
KEi . Since the package moves with constant velocity,
KE f
0 .
Note that the above result can also be obtained by the following reasoning:
Since the object has zero acceleration, the net (or resultant) force acting on it must be zero. The net work done is
Wnet
Fnet d
0 .
The work done by the conservative gravitational force is
Wgrav
PEg
mg y f
Wgrav
50 kg 9.80 m s2
yi
mg d sin
or
340 m sin 7.0°
2.0
104 J p
The normal force is perpendicular to the displacement. The work it does is
Wnormal
nd cos 90°
0
Since the package moves up the incline at constant speed, the net force parallel to the incline is zero.
Thus, F
0
fs
mg sin
0, or fs
mg sin .
Pages 5.45
Chapter 5
The work done by the friction force in moving the package distance d up the incline is
Wfriction
5.86
fk d
mg sin
50 kg 9.80 m s2 sin 7.0°
d
340 m
2.0
10 4 J
Each 5.00-m length of the cord will stretch 1.50 m when the tension in the cord equals the weight of the jumper
(that is, when F
s
mg ). Thus, the elongation in a cord of original length L when Fs
w
L
5.00 m
x
1.50 m
w will be
0.300 L
and the force constant for the cord of length L is
Fs
x
k
(a)
w
0.300 L
In the bungee-jump from the balloon, the daredevil drops yi
yf
55.0 m .
The stretch of the cord at the start of the jump is xi = 0, and that at the lowest point is x f
Since KEi
0
0 for the fall, conservation of mechanical energy gives
KE f
PEg
PEs
f
0
f
PEg
i
PEs
i
1
k x 2f
2
xi2
mg yi
giving
mg
1
2 0.300L
55.0 m
L
2
mg 55.0 m and
which reduces to
55.0 m
2
110 m L
L2
33.0 m L
or
L2
55.0 m
143 m L
55.0 m
2
0
and has solutions of
Pages 5.46
55.0 m
L
2
33.0 m L
yf
L.
Chapter 5
143 m
L
143 m
2
4 1 55.0 m
2
2 1
This yields
143 m
L
2
91.4 m and L
117 m or L
25.8 m
Only the L = 25.8 m solution is physically acceptable!
(b)
During the jump,
Fy
ma y
kx
mg
mg
x
0.300 L
ma y , or
mg
m ay
Thus,
x
0.300 L
ay
1 g
which has maximum value at x
ay
5.87
(a)
29.2 m
0.300 25.8 m
max
55.0 m
1 g
29.2 m .
L
27.1 m s2
2.77 g
While the car moves at constant speed, the tension in the cable is F = mg sin , and the power input is
mg sin ,
F
P
(b)
xmax
P
Fv
mgv sin
950 kg 9.80 m s2
, or
2.20 m s sin 30.0°
104 W
1.02
10.2 kW
While the car is accelerating, the tension in the cable is
Fa
mg sin
950 kg
ma
v
t
m g sin
9.80 m s2 sin 30.0°
2.20 m s
12.0 s
0
4.83
10 3 N
Maximum power input occurs the last instant of the acceleration phase. Thus,
Pages 5.47
Chapter 5
Pmax
(c)
Favmax
4.83
103 N 2.20 m s
10.6 kW
The work done by the motor in moving the car up the frictionless track is
Wnc
KE
PE
Wnc
950 kg
KE
f
PE
KE f
i
PEg
f
0
1
mv2f
2
mg L sin
or
5.88
(a)
1
2.20 m s
2
2
9.80 m s2
1 250 m sin 30.0°
5.82
10 6 J
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
PEg
f
KE
PEg
i
Choosing PEg = 0 at the level where the string attaches to the cart, this gives
0
mg
L cos
1
mv02
2
mg
L
or
v0
(b)
If L = 1.20 m and
v0
5.90
(a)
2gL 1
cos
= 35.0 , the result of part (a) gives
2 9.80 m s2 1.20 m 1
cos 35.0°
2.06 m s
Realize that, with the specified arrangement of springs, each spring supports one-fourth the weight of the
load (shelf plus trays). Thus, adding the weight (w = mg) of one tray to the load increases the tension in each
spring by F = mg/4. If this increase in tension causes an additional elongation in each spring equal to the
thickness of a tray, the upper surface of the stack of trays stays at a fixed level above the floor as trays are
added to or removed from the stack.
Pages 5.48
Chapter 5
(b)
If the thickness of a single tray is t, the force constant each spring should have to allow the fixed-level tray
dispenser to work properly is
F
x
k
mg 4
t
mg
4t
or
0.580 kg 9.80 m s2
k
4 0.450
10
2
316 N m
m
The length and width of a tray are unneeded pieces of data.
5.91
When the cyclist travels at constant speed, the magnitude of the forward static friction force on the drive wheel
equals that of the retarding air resistance force. Hence, the friction force is proportional to the square of the speed,
and her power output may be written as
P
kv2 v
fsv
kv3
where k is a proportionality constant.
If the heart rate R is proportional to the power output, then R
kP
k kv3
k kv3 where k is also a
proportionality constant.
The ratio of the heart rate R2 at speed v2 to the rate R1 at speed v1 is then
k kv32
k kv13
R2
R1
v2
v1
3
giving
v2
Thus, if R
v1
R2
R1
13
90.0 beats min at v
22.0 km h , the speed at which the rate would be is 136 beats/min is
Pages 5.49
Chapter 5
v
22.0 km h
13
136 beats min
90.0 beats min
25.2 km h
and the speed at which the rate would be 166 beats/min
v
5.92
(a)
22.0 km h
13
166 beats min
90.0 beats min
27.0 km h
The needle has maximum speed during the interval between when the spring returns to normal length and the
needle tip first contacts the skin. During this interval, the kinetic energy of the needle equals the original elastic
potential energy of the spring, or
vmax
(b)
xi
k
m
8.10
1
2
1
2
2
mvmax
10
2
kxi2 . This gives
375 N m
5.60 10 3 kg
m
21.0 m s
If F1 is the force the needle must overcome as it penetrates a thickness x1 of skin and soft tissue while F2 is
the force overcame while penetrating thickness x2 of organ material, application of the work–energy theorem
from the instant before skin contact until the instant before hitting the stop gives
Wnet
F1 x1
1
2
F2 x2
mv2f
1
2
2
mvmax
or
vf
f
2
vmax
21.0 m s
2 F1 x1
F2 x2
m
2
2
7.60 N 2.40
10
2
m
5.60
Pages 5.50
9.20 N 3.50
10
3
kg
10
2
m
16.1 m s