Download Chapter 5 Solutions

Chapter 5 Solutions
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
The net work done on the wheelbarrow is
Wnet  Wapplied  W friction   F cos 0  x   f cos180  x
force
  F  f  x   50.0 N  43 N   5.0 m   35 J
so choice (c) is the correct answer.
2.
In the absence of any air resistance, the work done by nonconservative forces is zero. The work–
energy theorem then states that KE f  PE f  KEi  PEi ,which becomes
1
1
m v2f  m gy f 
m vi2  m gyi
2
2
or
vf 

vi2  2 g yi  y f

Choosing the initial point to be where the skier leaves end of the jump and the final point where
he reaches maximum height, this yields
vf 
15.0

m s   2 9.80 m s2
2
   4.50 m 
 11.7 m s
making (a) the correct answer.
3.
The mass of the crate is


m  w g   40.0 N  9.80 m s2  4.08 kg
and we may write the work–energy theorem as
Wnet  Wfric  Wgrav  KE f  KEi
Since the crate starts from rest, KEi 
1
2
mvi2  0,
and we are left with
KE f  Wfric  Wgrav 
so
 fk cos 180 
 x   w cos 60.0  x
KE f   6.00 N   6.00 m    40.0 N  cos 60.0  6.00 m   36.0 J  120 J  84.0 J
and
vf 
2 KE f

m
2  84.0 J 
 6.42 m s
4.08 kg
making choice (d) the correct response.
4.
We assume the climber has negligible speed at both the beginning and the end of the climb. Then
KE f  KEi  0 , and the work done by the muscles is





Wnc  0  PE f  PEi  mg y f  yi   70.0 kg  9.80 m s2
  325 m   2.23  105 J
The average power delivered is
P 
Wnc
2.23  105 J

 39.1 W
t
95.0 min  60 s 1 min 
and the correct answer is choice (a).
5.
The net work needed to accelerate the object from  = 0 to  is
W1  KE f  KEi 
1
2
m v2 
1
2
m  0 
2
1
2
m v2
The work required to accelerate the object from speed  to speed 2 is
W2  KE f  KEi 
1
2
m  2v  
2
1
2
m v2 
1
2


m 4 v2  v2  3

1
2

m v2  3W1
Thus, the correct choice is (c).
6.
Assuming that the cabinet has negligible speed during the operation, all of the work Alex does is
used increasing the gravitational potential energy of the cabinet. However, in addition to
increasing the gravitational potential energy of the cabinet by the same amount as Alex did, John
must do work overcoming the friction between the cabinet and ramp. This means that the total
work done by John is greater than that done by Alex and the correct answer is (c).
7.
Since the rollers on the ramp used by David were frictionless, he did not do any work
overcoming nonconservative forces as he slid the block up the ramp. Neglecting any change in
kinetic energy of the block (either because the speed was constant or was essentially zero during
the lifting process), the work done by either Mark and David equals the increase in the
gravitational potential energy of the block as it is lifted from the ground to the truck bed. Because
they lift identical blocks through the same vertical distance, they do equal amounts of work and
the correct choice is (b).
8.
Once the athlete leaves the surface of the trampoline, only a conservative force (her weight) acts
on her. Therefore, her total mechanical energy is constant during her flight, or
KE f  PE f  KEi  PEi . Taking the y = 0 at the surface of the trampoline,
PEi  mgyi  0 . Also, her speed when she reaches maximum height is zero, or KE f  0 .
This leaves us with PE f  KEi , or m gymax 
vi2
 8.5 m s 

2g
2 9.80 m s2
1
2
m vi2 , which gives the maximum height as
2
ymax 


 3.7 m
making (c) the correct choice.
9.
KEcar 
1
2
mcar v2 
1
2
 mtruck 2  v2

1
2

1
2

mtruck v2  KEtruck 2 , so (b) is the correct
answer.
10.
The kinetic energy is proportional to the square of the speed of the particle. Thus, doubling the
speed will increase the kinetic energy by a factor of 4. This is seen from
KE f 
1
1
2
1

m v2f  m  2 vi   4  m vi2   4 KEi
2

2
2
and (a) is the correct response here.
11.
The work–energy theorem states that Wnet  KE f  KEi . Thus, if Wnet  0 , then
KE f  KEi or
1
2
m v2f 
1
2
m vi2 , which leads to the conclusion that the speed is unchanged
v
f

 vi . The velocity of the particle involves both magnitude (speed) and direction. The
work–energy theorem shows that the magnitude or speed is unchanged when Wnet  0 , but
makes no statement about the direction of the velocity. Therefore, choice (d) is correct but choice
(c) is not necessarily true.
12.
As the block falls freely, only the conservative gravitational force acts on it. Therefore,
mechanical energy is conserved, or. KE f  PE f  KEi  PEi Assuming that the block is
released from rest (KEi  0) , and taking y = 0 at ground level (PE f  0) , we have that
KE f  PEi
or
1
m v2f  mgyi
2
and
yi 
v2f
2g
Thus, to double the final speed, it is necessary to increase the initial height by a factor of four, or
the correct choice for this question is (e).
13.
If the car is to have uniform acceleration, a constant net force F must act on it. Since the
instantaneous power delivered to the car is P  F v , we see that maximum power is required just
as the car reaches its maximum speed. The correct answer is (b).
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 lower than the chosen reference level.
6.
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.
8.
(a) The effects are the same except for such features as having to overcome air resistance outside.
(b) The person must lift his body slightly with each step on the tilted treadmill. Thus, the effect is
that of running uphill.
10.
The kinetic energy is converted to internal energy within the brake pads of the car, the roadway,
and the tires.
12.
Work is actually
performed by the thigh
bone (the femur) on the
hips as the torso
moves upwards a
distance h. The force
on the torso Ftorso is
approximately the
same as the normal
force (since the legs
are relatively light and
are not moving
much), and the work
done by Ftorso
minus the work done by
gravity is equal
to the change in kinetic
energy of the
torso.
At full extension the
torso would
continue upwards,
leaving the legs
behind on the ground
(!), except that
the torso now does
work on the legs,
increasing their speed
(and decreasing
the torso speed) so that
both move
upwards together.
Note: An alternative way to think about problems that involve internal motions of an object is to
note that the net work done on an object is equal to the net force times the displacement of the
center of mass. Using this idea, the effect of throwing the arms upwards during the extension
phase is accounted for by noting that the position of the center of mass is higher on the body with
the arms extended, so that total displacement of the center of mass is greater.
14.
If a crate is located on the bed of a truck, and the truck accelerates, the friction force exerted on
the crate causes it to undergo the same acceleration as the truck, assuming that the crate doesn’t
slip. Another example is a car that accelerates because of the frictional forces between the road
surface and its tires. This force is in the direction of the motion of the car and produces an
increase in the car’s kinetic energy.