Download Chapter 6

Chapter 6
Answers to Questions
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12
Q13
Q14
Q15
Q16
Q17
Q18
Q19
Q20
Q21
Q22
The work done is greater for the block moving the greater distance. Work depends on force and distance,
not on mass.
No work has been done on the rock. All the force exerted by the man was countered by frictional forces
keeping the rock in place and thus the rock traversed no distance.
a. Yes, assuming there is friction between the block and floor. If there were no friction, once set in motion
no force would be required to keep it moving at constant velocity and no work. With friction present
work is required to overcome the negative work done by friction.
b. The force involved in the work is the component of force along the line of the motion.
Yes. There is a frictional force acting on a body over a distance. The friction does an amount of negative
work equal to the positive work done by the string. However, this energy is transferred out of the system in
the form of heat dissipated into the atmosphere.
No. The normal force has no component along the line of motion and cannot speed up or slow down the
body.
No. That force is perpendicular to the direction of the motion.
Normal force from floor, gravity, friction from air, friction from floor. The frictional forces do work, because
they are parallel to the direction of motion. In this case, they do negative work, removing kinetic energy from
him.
The work done by the woman is equal to the work done by the rope on the crate. Work is determined by the
amount of force times the distance traveled,
which for the rope and the crate is equal. Although the force on the rope is one-fourth the weight (force) of
the crate, the rope traveled four times the distance traveled by the crate.
The work done by the person can never be less than the work done by the lever on the rock. If there are no
dissipative forces, they will be equal. This is a consequence of the conservation of energy.
Yes. A simple machine’s function by definition is to amplify the effect of the energy transferred.
In the absence of friction, all of the work done by the force exerted by the boy doing the pushing goes into
the change in kinetic energy. (Work is being converted to energy of motion.)
No. The net work done on the block (net energy transferred to the block changing its motion) is equal to the
change of its kinetic energy. The work done by the tension in the string is larger to compensate for the
energy dissipated by friction.
Not necessarily. Let us distinguish two cases. If there is a force component along the line of motion, the
force could act opposite to the motion and cause a decrease in kinetic energy. If the force is at right angles
to a body moving in a circular path, this centripetal force does no work and cannot change the kinetic
energy.
No. The work done on the faster ball is four times the work done on the slower ball. The work goes into
2
kinetic energy and kinetic energy is proportional to v .
A force equal to the weight of the box has acted through a distance. The work has gone into increasing the
potential energy of the block-earth system.
NO! If the kinetic energy doesn’t increase, then the speed is constant, thus the acceleration is zero. Hence,
the net force is zero.
Yes. A system that has all its energy in the form of potential energy is such a system. Potential energy
depends on position relative to a reference point.
Yes. The weight of the crate has been lifted slightly. If it is released, it will fall back and convert the
potential energy into kinetic energy.
The majority of Earth’s mass is sufficiently below both so that acceleration due to gravity is about the same
for both of them. Potential energy depends on an arbitrary zero point, not an absolute, like sea level. Thus
the ball with the 20m drop into a well has more potential energy than the 10m drop to Earth’s surface.
The work in cocking the bow and arrow has been transferred into elastic potential energy of the bow.
Ouch! She has added energy in the form of kinetic energy to the potential energy of the system. Unless
there is enough friction in the apparatus to counter this, it’s time to call for an ambulance!
a. The work in raising the pendulum bob has gone into gravitational potential energy.
b. The kinetic energy is greatest where the potential energy is the least—at the bottom of the arc where
the height is smallest.
c. The potential energy is the greatest when the kinetic energy is the least—at each end of the motion
where the bob comes to rest instantaneously as it reaches the highest point in its motion.
Q23
Q24
Q25
Q26
Q27
Q28
Q29
Q30
Q31
Q32
Q33
Q34
Q35
Q36
Because the bob is half way down its arc, the total energy consists of ½ of the original potential energy and
½ of the original potential energy converted to kinetic energy.
Since a pendulum experiences air resistance and friction of moving parts in contact, it will lose mechanical
energy and eventually stop swinging.
a. The energy from burning the fuel in the car is transformed into kinetic energy for the car to move. The
kinetic energy of the car is transformed into heat due to the friction with the road when it burns rubber.
Some of the kinetic energy is used to burn the rubber also.
b. Yes, total energy is always conserved, but some mechanical energy has been converted to heat, so total
mechanical energy has not been conserved.
a. The chemical energy of the fuel is transformed into the kinetic energy of the car. Some of the kinetic
energy is transformed into heat energy due to friction between the SUV and the road and the air, and
the friction between different parts of the SUV itself.
b. Mechanical energy is NOT conserved in this situation because some of the kinetic energy is
transformed into heat due to frictional forces.
c. Energy of all forms is conserved.
a. Yes. The energy from burning the oil goes into heating the air (and thus our hands).
b.
We are using a high grade form of energy. We are wasting a lot of the heat generated from an
expensive source of energy just to heat the atmosphere, and we are also polluting the atmosphere.
As the bird carries the clam upward, the potential energy of the clam (not to mention the bird) increases.
Work is done to carry the clam aloft. When the bird drops the clam, the clam’s potential energy is converted
to kinetic energy with some of the energy being dissipated as heat by the frictional force of air resistance.
When the clam hits the rock all of the kinetic energy is dissipated by the impact (an example of a mostly
inelastic collision), but some goes into the kinetic energy of the various shell shards that fly off so that the
bird can now enjoy its lunch.
No. The vaulter also adds kinetic energy of running. The elastic PE of the flexible pole and some of the
kinetic energy becomes gravitational PE as the vaulter ascends.
Not necessarily. The height that each reaches will depend on the vaulter's strength and ability to work his
body as he jumps, and also his skill at converting all of the kinetic energy into potential energy.But if two
vaulters are equally skilled, a faster one will generally be able to reach a greater height.
a. The work in stretching the spring goes into elastic potential energy.
b. The potential energy is the greatest when the kinetic energy is the least—at each end of the oscillation
when the body comes to rest instantaneously and the spring is compressed or stretched by the
maximum amount.
c. The kinetic energy is greatest as the mass moves through the equilibrium point; for here the potential
energy has all been transferred to kinetic energy.
The energy of the system is a combination of kinetic and potential energies. (Since the potential energy
depends upon the distance from the center squared, the potential energy at the halfway point is just 1/4 of
that at the extreme point. The kinetic energy is thus 3/4 of the total energy at the halfway point.)
The cocking of the dart gun adds elastic potential energy to the system. The dart is then released, so the
elastic potential energy is converted to kinetic energy for the time the dart is in contact with the spring. Then
the kinetic energy of the dart is converted into potential energy, slowing the dart as it rises toward the ceiling.
Also, there is the transformation of kinetic energy into heat due to the frictional force of the air.
Increased. There are two potential energy contributions. As the mass is lowered the gravitational potential
energy decreases while the elastic potential energy of the spring increases. The force stretching the spring
is greater than the weight of the object so there is a net increase in potential energy.
Yes, if the initial kinetic energy given by the push is greater than the additional potential energy of the body
when at the hump.
Yes. When a car accelerates, the force of static friction between the tires (if it were not static friction, the
Work
tires would be sliding!) and the road actually increases the mechanical energy of the car. Q37
output will be less than work input because some of the energy from the input must go to counter the
negative work of the rusted pulley surfaces.
Answers to Exercises
E1
E2
E3
E4
75 J
40 N
5m
a. 380 J
b. 100 J
c. 280 J
E5
E6
E7
E8
E9
E10
E11
E12
E13
E14
E15
E16
E17
E18
E19
a. 160 J
b. 0 (zero) J
c. 160 J
a. 600 J
b. 600 J
a. 125 J
b. 125 J
a. 98 J
b. 98 J
a. 0.6 J
a. 40 J
b. 2,000 N/m
2
(Using g = 9.8 m/s ) Accelerating the rock. The acceleration from rest requires 100 J versus 78.4 J to lift the
rock.
a. 1.6 J
b. 0.82 m
50 J
a. 6860 J
b. 8560 J
c. 8560 J
520,000 J
a. 264.6 kJ
b. 119.1 kJ
c. 105.5 kJ
a. 32 J
b. 32 J
4 Hz
0.125 seconds
Answers to Synthesis Problems
SP1
SP2
SP3
SP4
SP5
SP6
a.
b.
c.
d.
7.5 J
4.5 J
4.5 J The net force (3N) creates net work, which increases the kinetic energy by F x d.
3/5 of it goes into changing the KE. The rest is converted to thermal energy. Yes. It can all be
accounted for. Energy is always conserved for all kinds considered.
e. 4.5J, 6 m/s
2
a. 0.5 m/s
b. 4 m
c. 200 J
d. 2 m/s
e. 200 J; Equal to the amount of work input as calculated in part c.
a. 48 J
b. 48 J Assuming that it is fired horizontally.
c. 43.8 m/s
d. No. Internal friction in the band, air resistance, and aiming upwards will reduce the maximum possible
kinetic energy. Yes. As for the strap, because it moves, it is gaining kinetic energy.
a. 1.73J
b. 4.16 m/s. Maximum velocity occurs as the mass moves through the equilibrium position.
c. 0.432 J, 1.30 J, 3.60 m/s
d. 0.87 (About 87% of maximum velocity)
a. Yes. The difference between the potential energy at the first point and the second point is 3920J,
whereas the loss of energy due to friction is only 2000J.,
b. 34.9 m. This additional height would allow the body to reach this point just as it has lost all of its kinetic
energy.
a. 100 J (W = Fd = 100N x 1m = 100Nm=100J)
b. 100 J (W = Fd = 50N x 2m = 100Nm = 100J)
c. The same amount of work is required for both situations.
d. Lifting straight up requires more force.
e. The force is applied over a longer distance when using the ramp.
∆PE = mg∆h is the same for both since they reach the same final height. Note that this is consistent
f.
with the answer to part c, since the change in the gravitational potential energy should be equal to the
work done.
g.
Since you need only apply half the force when using the ramp, you can conserve your strength when
using the ramp.
Chapter 7
Answers to Questions
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12
Q13
Q14
Q15
Q16
Q17
Q18
Yes. Impulse is the product of the average force times the time the force acts.
The first force is the larger force because impulse is equal to the product of force and time. If the second
time is twice that of the first, then the first force must be twice that of the second force.
Momentum is mass times velocity, so a fast baseball could possess the same momentum as a slow bowling
ball.
No. Impulse and force are not the same thing. Impulse depends on force, but it also depends on the length
of the time interval the (average) force was applied.
No. Impulse and momentum are not the same thing. Impulse is equal to the change in momentum, either its
magnitude or its direction or both.
a. Yes. Momentum is a vector pointing in the direction of the velocity, so a change in direction implies a
change in momentum.
b. Yes. The force exerted by the wall on the ball first slows the ball down and continues to act on the ball
as it starts to speed up in the opposite direction until it clears the wall. The impulse is the average force
on the ball times the time it acts.
The change in momentum of the baseball depends on the impulse. By following through, the bat can remain
in contact with the ball longer, yielding a larger impulse.
A padded dashboard gives a longer impact time; the change of momentum (impulse) takes longer thus
decreasing the average force of the dash against the rider’s face/body. Average force is decreased because
the force is given a longer time to vary. Force is the change of momentum over time and a longer time gives
a smaller force for the same change of momentum.
The airbag lengthens the amount of time and distance to stop the occupant. This reduces the amount of
force necessary to stop the occupant.
A hard, rapidly expanding airbag is somewhat equivalent to a hard dashboard and does not significantly
lengthen the amount of impact time.
Yes. This will result in a longer impact time and the change of momentum will increase, thus decreasing the
average force of the ball against the hand.
More likely. If you move your hand forward, you will reduce the time it takes to bring the egg to a full stop. If
the time decreases, the force increases, according to the definition of impulse. Therefore you are more likely
to break the egg if you do anything to shorten the time it takes to bring it to a stop.
The truck will require the larger impulse to stop it. Impulse is the change of momentum and though both the
bicycle and the truck are going from the same velocity to zero, the truck’s larger mass gives it the greater
momentum change.
Momentum is conserved for any system when the net force acting on the system is equal to zero.
No. Momentum is not conserved since there is a net force on the ball, which is the component of weight
along the incline.
No, but the total momentum of the two-body system remains constant. An object exerts an impulse on a
second object changing the momentum of the second object. The reaction of the second object provides an
impulse on the first object changing its momentum. Overall, since the action and reaction forces are equal
and opposite and the time each force acts is the same for both, the change in momentum of the second will
be equal and opposite to the change in momentum of the first leaving the total momentum of the system
unchanged.
Conservation of momentum for a single body follows from Newton's first law. For a system of interacting
bodies, Newton's third law justifies the fact that there is no net force on the system as a whole, which is the
condition for the total momentum to be conserved.
a. The magnitude of the average force on each vehicle will be the same.
b. Since the force on each vehicle and the time the force acts are the same, each has the same impulse in
magnitude.
c. Both vehicles experience the same magnitude of change of momentum, since each experiences the
same impulse.
d.
Q19
Q20
Q21
Q22
Q23
Q24
Q25
Q26
Q27
Q28
Q29
Q30
Q31
Q32
Q33
Q34
Q35
Q36
Q37
The compact car will suffer the greater acceleration. The acceleration is force divided by mass. Both
vehicles will be acted upon by the same magnitude of force, but the mass of the compact car is much
less than that of the truck.
It is possible if the lighter defensive’s momentum is greater than the heavier fullback’s.
Zero. Because of the third law, the force on each skater is equal and opposite in direction. The momentum
gained by one skater will be equal and opposite that gained by the other.
If free to move, both shotguns will possess the same momentum upon recoiling. The lighter one will recoil
faster.
Yes. The total momentum of the boat + gun before firing is zero. After firing the shell, the forward
momentum of the shell plus the backward momentum of the boat (with gun) must still be zero; hence, the
recoil of the boat. Since the boat can be very massive compared to the shell, the recoil speed of the boat
may be quite small.
The rocket moves forward because the exhaust material ejected backwards relative to the motion of the
rocket exerts a reaction force on the rocket.
Throw the bag of oranges away thus propelling you in the opposite direction. This would be equivalent to
“pushing off’’ against the bag of oranges. One might also toss away the oranges singly for a more controlled
mode of travel.
She should remove a wrench from her tool belt and throw it in the opposite direction to the location of the
space shuttle. By conservation of momentum, she will then gain an equal amount of recoil momentum in the
direction of the shuttle.
a. At fan: to the right. At sail: to the left.
b. On fan: to the left. On sail: to the right.
c. If he doesn't have the sail rolled up (furled), he won't go anywhere. The fan on the boat blowing air
against an open sail is an example of internal force. Internal forces do not impart acceleration to a body.
With the sail furled (rolled up) he can move opposite the way the fan is blowing. The fan will then be
blowing against an external body (the atmosphere).
The skateboard will slow down. By adding the mass of the skateboarder to that of the skateboard, the total
mass of the system will increase. By conservation of momentum, if the mass increases, the velocity must
decrease in order for the momentum mv to remain constant.
Momentum is conserved, therefore momentum after the collision must equal momentum before the collision
(due to the lack of external forces in this problem). Because the mass of the moving system is increased by
the addition of the second car, the velocity will decrease.
The coupling of the two cars after collision satisfies the conditions for a perfectly inelastic collision.
No. Momentum is conserved in all collisions, elastic or inelastic, as long as there is no net external force.
No. There is a loss of kinetic energy in the collision, so it is inelastic.
a. No. The direction of the ball’s momentum vector was changed by an external force (the wall).
b. Define the system as the ball, the wall anchored to Earth; then yes, momentum of the system is
conserved.
Yes. The collision is elastic. There has been no change in the kinetic energy of the system.
The resultant momentum will point in the direction of the larger momentum, p1, and will have a magnitude
(p1 - p2).
o
No. In this problem the final momentum will be in a direction making a 45 degree angle with respect to
each of the initial momentum vectors.
If the velocities had been equal, the direction of motion after the perfectly inelastic collision would have been
o
at a 45 angle with respect to the direction of A. Since the angle is smaller, A must have had a larger
momentum and thus was traveling faster.
The momentum of the truck is twice that of the car, so the final direction of the coupled vehicles is closer to
the original direction of the truck than that of the car.
Q38
before
after
Answers to Exercises
E1
E2
E3
E4
E5
E6
E7
E8
E9
E10
E11
E12
E13
E14
E15
E16
a. 12.0 Ns
b. 12.0 kg m/s
3
P = 48.6 x10 Ns
The bowling ball has the larger momentum.
a. 9 Ns
b. 9 kg m/s
5.4 Ns
a. 24 Ns
b. 160 N
a. 1380 Ns
b. 3450 N
a. -5.0 Ns
b. -5.0 Ns
a. -7.5 kg m/s
b. -7.5 Ns
a. Pfb = 350 kg m/s; Pdb = -480 kg m/s
b. 130 kg m/s (east)
c. east
a. Zero
b. -7.5 m/s
a. 4.2 kg m/s
b. 3.5 m/s
4
2.0 x 10 kg m/s in the forward direction.
5
a. 1.08 x 10 kg m/s
b. 3.6 m/s
6
a. 2.25 x 10 J
5
b. 4.5 x 10 J
c. Kinetic energy is not conserved; some is lost to
frictional energy during coupling.
4
4
a. Pt = 4.0 x 10 kg m/s north; Pc = 2.7 x 10 kg m/s south.
4
b. 1.3 x 10 kg m/s, north
E17
North
a.
Ptruck
Pcar
4
Scale: 0.5 cm = 10 kg m/s
b.
Ptotal
Pcar
Ptruck
4
Scale: 0.5 cm = 10 kg m/s
E18
a.
4
Scale: 1 cm = 10 kg m/s
b.
4
5.0 x 10 kg m/s
Answers to Synthesis Problems
SP1
SP2
a.
b.
c.
d.
a.
b.
c.
SP3
a.
b.
c.
SP4
a.
-12 kg m/s
Yes. Momentum changed direction.
12 Ns
300 N
2.07 m/s
2.49 kg m/s
Technically, No. The bullet is still moving, so it has some momentum. However, it is negligible when
compared to that of the block.
Case B
Case B
For the ball alone, momentum is not conserved because direction changes. For the system of ball plus
wall attached to the earth, momentum is conserved in both cases.
-1620 kg m/s
c.
SP5
a.
b.
c.
d.
e.
Wearing a seatbelt distributes the stopping force along the portions on the body restrained by the belt,
usually muscular areas of the body. Hitting the dashboard places all the force at the small contact point
where the head hits the windshield. The time of action will be slightly longer in the seatbelt scenario
because of the action of the seatbelt stop action and the possible stretching of the belt. The major
advantage to a seatbelt, though, is spreading the force out over a larger area, not in extending the time
over which the force is acting.
30,000 kg m/s, South
5 m/s South
5
9.75 x 10 J
4
7.5 x 10 J
No. The collision is inelastic; the vehicles stuck together and kinetic energy was not conserved.
Chapter 8
Answers to Questions
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12
Q13
Q14
Q15
Q16
Q17
Q18
Q19
Q20
Q21
2
Rotational velocity is rotational displacement divided by time, but m/s is linear velocity and rev/s is rotational
acceleration; rad/s and rev/m would thus be appropriate units.
2
2
Rad/s and rev/m are inappropriate. Rad/s is angular velocity and rev/m is meaningless.
Yes.Any body whose rotational velocity is changing is exhibiting rotational acceleration.
Yes. Any change in rotational velocity is a rotational acceleration, even if it is caused by a torque in the
direction opposing the rotation.
Yes. The merry-go-round is a rigidly rotating body. All parts along a given line from center to edge are
rotating at exactly the same rotational velocity.
No. Since the relation for the distance along the arc of a circle is equal to the radius times the rotational
displacement in radians, the inner arc is smaller than the outer arc. Therefore the outer child’s linear velocity
is greater than that of the inner child.
No. Constant acceleration results in constantly changing velocity.
The ball experiences both linear and angular acceleration. The ball travels one circumference along the
incline during one revolution, as measured along the incline. The magnitude of the linear velocity is equal to
the radius of the ball multiplied by the magnitude of the angular velocity. The direction of the linear velocity
is perpendicular to the direction of the angular velocity.
The force applied at the end of the wrench will provide the greater torque because the length of the lever
arm associated with that force is greater.
F2. F1 has no component perpendicular to the radius. .
F1 provides the larger torque. F2 has a smaller component perpendicular to the radius.
Yes, if the respective distances (lever arms) from the fulcrum are chosen properly. The ratio of the distances
will be in the inverse ratio of the weights.
Yes, if they do not act along a common line. A simple rod with a force upward on one end, and downward on
the other end. There are many other arrangements that yield a net force of zero, yet have a net torque.
To get a magnification of force, the fulcrum should be placed closer to the rock, so that the lever arm for the
applied force is greater than the lever arm for the opposing weight.
No. A net torque of zero (i.e., balance) at this point indicates that the center of gravity for this pencil is
located at the same place the fulcrum is positioned.
The plank can be pushed to its center point, for that is where the center of gravity is located. Beyond that
point less of the plank’s mass will be supported by the platform than will be hanging over the edge. This
condition will result in a net torque downward.
No. The shape of the wire and distribution of its mass are such that the point about which its own weight
exerts no net torque is not on the wire. Therefore the center of gravity will not lie on the wire itself.
No. A net torque produces an angular acceleration.
The tall crate will be the easiest to tip over because the moment arm is the greatest when the forward edge
is the axis of rotation.
Body A will have a smaller rotational inertia, so that for a given torque it will acquire a larger angular
acceleration than if the torque were applied to B.
Yes, if the same mass is distributed at different distances from the axis of rotation in the two cases.
Q22
Q23
Q24
Q25
Q26
Q27
Q28
Q29
Q30
Q31
Q32
Q33
Q34
Q35
Q36
Q37
Q38
Yes. Extending one's arms will increase the rotational inertia compared to the case of the arms at one's side.
The hollow sphere has the greater rotational inertia because its mass is distributed farther from the axis of
rotation.
No. Angular momentum is only conserved when there is no net torque.
No. The distribution of mass gives a different rotational inertia. Therefore the angular momentum will also be
different.
It will decrease. After the child is near the edge, the rotational inertia of the system will be greater; so to
conserve angular momentum, the final rotational velocity must be lower.
The increased mass will slow it down.
Yes. He can change his rotational inertia by changing the extension of his arms. His rotational velocity changes
according to conservation of angular momentum.
The rotational velocity will increase due to the decrease in rotational inertia.
The direction of the angular momentum is perpendicular to the plane of the wheel at any instant, so in turning a
o
corner the direction will change by 90 .
No. Since the yo-yo is already spinning, it continues spinning in the same direction, and thus the angular
momentum vector is still pointing in the same direction.
A sleeping yo-yo requires the string to slip on the axis of the yo-yo. If it is tied tightly, it will wind back up after it
is dropped.
By definition the counterclockwise direction as viewed from above results in the rotational velocity vector
pointing upward. Therefore the angular momentum vector is pointing upward because it is parallel to the
velocity vector, and since the skater is rotating counterclockwise the velocity vector is pointing upward.
a.Away from the observer.
b. No. During the fall gravity provided the external torque acting about the pivot point .
Due to the conservation of angular momentum, a spinning top maintains its orientation until frictional forces
slow the top significantly. As the spin slows, the top begins to precess until gravitational torque is dominant
enough to topple the toy over.
Yes. The size of the torque transmitted to the rear wheel depends on which of the several sprockets is
engaged with the chain because the moment arm changes.
When the pedal is in the forward position and we push perpendicularly to the shaft, we will exert the
maximum torque. The magnitude and the direction of the applied force are most effective in this position.
A lower gear. For a lower gear, we need to move the chain to a larger sprocket on the rear wheel. This will
transmit a larger torque to the rear wheel at the expense of turning the wheel through a smaller angle and
moving the bicycle through a smaller distance for each turn of the crank.
Answers to Exercises
E1
E2
E3
E4
E5
E6
E7
E8
E9
E10
E11
E12
E13
E14
E15
E16
a. 0.17 rev/s
b. 1.05 rad/s
a. 0.75 rev/s
b. 3.75 revolutions
a. 31.4 rad
b. 7.85 rad/s
2
a. 0.3 rad/s
a. 4.8 rev/s
b. 9.6 rev
2
-0.42 rev/s
a. 1 rev/s
b. 2.5 rev
a. 12 N•m
b. 6 N•m
15 cm
16 N
a. 96 N•m
b. -60 N•m
c. 36 N•m
2
5 rad/s
16.8 N•m
a. 45 N•m
2
b. 15 kg•m
2
0.10 kg•m
2
a. 0.20 kg•m
2
E17
E18
b. 0.60 kg•m /s
2
a. 0.18 kg•m
2
b. 3.6 kg•m /s
6.28 rad/s = 60 rpm
Answers to Synthesis Problems
SP1
SP2
a.
b.
c.
d.
a.
b.
c.
SP3
a.
b.
c.
d.
SP4
a.
b.
c.
132 N•m; Directed along the merry-go-round axis.
2
0.147 rad/s
2.2 rad/s
-2
2
-1.3x10 rad/s ; 165 s after pushing is stopped.
80 N•m
0.53 m
He could have somebody else be prepared to hold down the other end if the plank started to tip. Or do it
in winter when the lake is frozen over. Or during a drought when the water levels are down. Or have a
boat waiting. Or…
2
2
960 kg•m ; 2460 kg•m
2
1560 kg•m
1.89 rad/s
Yes. During the time the children are moving, friction between their feet and the merry-go-round produces
the accelerating torque.
2
62.8 kg•m /s upward from plane of wheel.
20.9 rad/s (3.33 rev/s) in the direction of the original angular velocity.
The student exerts forces on the handles when he flips the wheel, producing the torque on the wheel,
which then produces an equal and opposite torque on him.
Chapter 9
Answers to Questions
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12
Yes. A 100 lb woman can exert pressure high enough to dent an ordinary floor tile if she puts her weight on
narrow spike heels. Pressure is force/area, so a moderate force on a small area results in a large pressure.
As pressure is defined as force per unit area, the unit of pressure in the foot-pound system would be given
2
as pounds per square foot (lb/ft ).
The pressure will be greater for the piston with small area since pressure is force/area.
The quarter. The pressure on each coin is the same, but since the quarter has a larger area, it will
experience the greater force.
The area of contact of the bicycle tire is much smaller than that of an auto tire so that even though the car is
much heavier, the bicycle tire requires a higher air pressure to provide the necessary force.
a. The larger piston experiences the greater force since F = p x A.
b. The large-area piston moves a smaller distance than does the small-area piston. The work must be the
same for each, so because of the greater force on the large-area piston, it will move a smaller distance.
No. The output piston exerts a greater force because its surface area is greater, but the pressure on each is
the same.
No. The pressure of any air in the top end would reduce the barometer reading from the true value of
atmospheric pressure.
Yes, except it would have to be a tube 13.6 times as long as that for a mercury barometer—over 10 m long!
That would be awkward to read and would not fit indoors.
a. mm hg
b. gauge pressure. The gauge pressure is simply the difference between that being measured and the
atmospheric pressure.
A giraffe. The head of a giraffe is often considerably higher that its heart, so a higher pressure is needed to
pump blood to the head. A dog's head is seldom much higher than its heart.
The level would decrease. The barometer is measuring the weight/area of air above it, so as we climb the
mountain the pressure decreases.
Q13
Q14
Q15
Q16
Q17
Q18
Q19
Q20
Q21
Q22
Q23
Q24
Q25
Q26
Q27
Q28
Q29
Q30
Q31
Q32
Q33
Q34
The balloon will contract. As the atmospheric pressure outside the balloon increases, the air in the balloon
will contract so that the inside pressure equals the outside pressure.
Atmospheric pressure changes with altitude. When traveling in mountains, the atmospheric pressure will
change faster than the body’s ability to change the pressure in the inner ear; hence, the popping.
The pressure decreases. If the volume is slowly increased, heat can flow in to keep the temperature
constant giving a Boyle's Law situation. The pressure decreases as the volume increases.
A drop in atmospheric pressure generally accompanies storms. The balloon will expand, assuming the
temperature is constant.
3
Yes, as long as its density is less than that of mercury, 13.6 g/cm .
The pressure pushing on the bottom is greater than the pressure pushing on the top. The pressure in a
liquid increases with increasing depth.
Yes, if it is shaped so that it displaces a sufficient volume of water equal to its weight. After all, steel ships
do float.
a. Equal. This is an equilibrium situation.
b. For the body to float, the volume of fluid displaced must be less than the volume of the block.
The water level will increase by the amount of water the bird would displace if it were in the water.
Yes. The density of salt water is slightly higher than that of fresh water, so it can exert a larger buoyant
force for a given displaced volume of water.
Water level in pool drops but the boat will ride higher. The original level of the boat took into account a
displacement of water equal to the weight of the anchor. In the water, the anchor does not displace its
weight in water, which is why it sinks.
It will by partly submerged. The object will sink until the weight of the displaced water is equal to the weight
of the object. If the density of the water is greater than the object, it will float partially submerged.
The velocity of the water decreases. For steady flow, the product of velocity times the area is constant.
Along a streamline, pressure is lower where the velocity is higher. As the water falls, its velocity increases.
The lower pressure in the center of the stream allows the pressure of the atmosphere to constrict it
somewhat.
No. The liquid with low viscosity will flow more rapidly. Viscosity is a frictional effect retarding the flow.
No. The transition from laminar to turbulent flow occurs as one goes from low to higher velocities.
The upward flow of cigarette smoke is accelerated by a buoyant force. Its increased speed and its low
viscosity cause the streamlines to become chaotic.
They come together. The higher air velocity between them means that the pressure will be reduced in that
region, according to Bernoulli's Principle, and atmospheric pressure above and below pushes them together.
The door will swing open. It swings toward the plane of the gust of wind that is flowing over its open surface.
Pressure is lowered by the moving air on that side of the open door.
The air will be moving fastest in the center so the air pressure is lowest at that point.
The ball is spinning so that the bottom of the ball is moving in the same direction as the linear velocity of the
ball. We would call this “back-spin” and it causes the ball to rise. Thus the batter would see it rotating
counterclockwise.
Yes. There is a deflecting force on the ball. The degree of curvature depends on the rate of spin on the ball
and the roughness of the surface of the ball.
Answers to Exercises
E1
E2
E3
E4
E5
E6
E7
E8
E9
E10
E11
E12
80 Pa
375 psi
1.25 psi
90 N
a. 400 kPa
4
b. 8.0 x 10 N
120 N
2750 Pa
3
0.04 m
26.67 kPa
2.94 N
3
500 kg/m
a. 4500 kg
4
E13
E14
E15
E16
b. 4.41 x 10 N
3
3.92 x 10 N
2.0 m/s
1/2
6000 Pa
Answers to Synthesis Problems
SP1
SP2
SP3
SP4
SP5
a.
b.
c.
d.
a.
b.
c.
d.
e.
a.
b.
c.
d.
e.
a.
b.
c.
d.
a.
b.
c.
2
2
3.14 cm , 490.87 cm
156.25:1
4
1.372 x 10 N
87.8 N
3
1.5 m
3
1.5 x 10 kg
4
1.47 x 10 N
29.4 kPa
0.291 atmosphere
-5
3
2.7 x 10 m
0.21 kg
2.06 N
0.265 N
1.80 N
11760 N
11760 N
3
1.20 m
0.267 m
2
2
50.26 cm , 19.6 cm
3.85 m/s
The pressure in the narrow portion where the velocity is greater will be less than the pressure in the
wider portion, according to Bernoulli's Principle applied to a horizontal streamline.