# TRUE This is the definition of power

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```					MOMENTUM AND ENERGY REVIEW 2

Complex Analysis and Problem-Solving: The following two problems involve higher order
thinking, but you have all the tools to solve these problems, draw diagrams and analyze the
situation.

QUESTION: Suzie Lovtaski has a mass of 49.7 kg. She is at rest on top of a hill with a height of
92.6 m and an incline angle of 19.2 degrees. She coasts down the hill to the bottom and
eventually comes to a stop; she never uses her poles to apply a force. The coefficient of friction
is 0.0873 along the hill and 0.527 along the horizontal surface at the bottom. What total distance
will Suzie coast (include both incline and level surface)? PSYW

This problem is very similar to question #39 and can be treated in much the same way. There is a
non-conservative force - friction - doing work upon the skier. This force will alter the total
mechanical energy of the skier. The equation to be used is

KEi + PEi + Wnc = KEf + PEf

If we designate the level area at the bottom of the slope as the zero level of potential energy, then
PEf is 0 J. Since Suzie eventually stops (due to the effect of friction along the level area), the KEf
is 0 J. And since Suzie starts from rest, KEi = 0 J. So the above equation becomes

PEi + Wnc = 0

The Wnc term has two parts; there is friction doing along the inclined plane and friction doing
work along the level surface. Since these two sections of the motion have different normal forces
and friction coefficients (and therefore friction forces), they will have to be treated separately.
The graphic below depicts the free-body diagrams and the means by which the friction force can
be determined.

By substituting values of mu and mass and g and theta into the above equations, one finds that
the friction values are
On Incline                         On Level Surface

Ffrict = 40.2 N                          Ffrict = 257 N

These forces act upon the skier over different distances. In the case of the inclined plane, the
distance (d) can be computed from the given incline angle and the initial height. The relationship
is depicted in the diagram below. The sine function is used to relate the angle to the initial height
and the distance along the incline. In the case of the level surface, the distance is the unknown
quantity (x) which this problem calls for.

The distance d along the incline is

d = hi / sin(theta) = 92.6 m / sin(19.2 deg) = 282 m

Now substitutions can be made into the work-energy equation and algebraic manipulation can be
performed to solve for x:

PEi + Wnc = 0

PEi + Wincline + Wlevel = 0

(49.7 kg)•(9.8 m/s)•(92.6 m) + (40.2 N)•(282 m)•cos(180 deg) + (257 N)•(x)•cos(180 deg) = 0 J

45102 J - 11307 J - 257 x = 0 J

33795 J = 257 x

132 m = x

QUESTION: Pete Zaria applies a 11.9-Newton force to a 1.49-kg mug of root beer in order to
accelerate it from rest over a distance of 1.42-m. Once released, how far will the mug slide along
the counter top if the coefficient of friction is 0.728? PSYW

Here is an instance in which there is only horizontal motion and a non-conservative force is
doing work to change the kinetic energy of the mug. The entire motion will have to be divided
into two parts - the part in which both friction and applied force act upon the mug (Pete is
pushing it from position A to position B) and the part in which the mug is slowing down under
the sole influence of friction (from position B to position C). The work-energy theorem (W =
Delta KE) will be used to analyze each part.

From position A to position B, Pete is pushing the mug and the net force upon the mug is simply
the applied force minus the friction force (see free-body diagram below). From position B to
position C, friction is the unbalanced force and acts as the net force (see free-body diagram
below).

The force of friction is related to the normal force (= m•g) and the coefficient of friction (0.728).
The Ffrict is

Ffrict = mu•Fnorm = (0.728) • (1.49 kg) • (9.8 m/s/s) = 10.6 N

This means that the net force from A to B is 11.9 N - 10.6 N = 1.27 N. The net force from B to C
is 10.6 N.

From A to B, the work done equals the kinetic energy change. So the kinetic energy at position B
is

KEB = W = F • d • cos(theta) = (1.27 N) • (1.42 m) • cos(0 deg) = 1.80 J

From B to C, the mug will lose this same amount of kinetic energy as friction works upon it to
bring it to a stop. So the work done from B to C is -1.80 J.

W = Change in KE

F • d • cos(theta) = -1.80 J
(10.6 N) • (d) • cos(180 deg) = -1.80 J

- (10.6 N) • (d) = -1.80 J

d = 0.170 m
ENERGY REVIEW
Part A: Forced Choice Questions

1. Which of the following statements are true about work? Include all that apply.

a.   Work is a form of energy.
b.   A Watt is the standard metric unit of work.
c.   Units of work would be equivalent to a Newton times a meter.
d.   A kg•m2/s2 would be a unit of work.
e.   Work is a time-based quantity; it is dependent upon how fast a force displaces an object.
f.   Superman applies a force on a truck to prevent it from moving down a hill. This is an
example of work being done.
g.   An upward force is applied to a bucket as it is carried 20 m across the yard. This is an
example of work being done.
h.   A force is applied by a chain to a roller coaster car to carry it up the hill of the first drop
of the Shockwave ride. This is an example of work being done.
i.   The force of friction acts upon a softball player as she makes a headfirst dive into third
base. This is an example of work being done.
j.   An eraser is tied to a string; a person holds the string and applies a tension force as the
eraser is moved in a circle at constant speed. This is an example of work being done.
k.   A force acts upon an object to push the object along a surface at constant speed. By itself,
this force must NOT be doing any work upon the object.
l.   A force acts upon an object at a 90-degree angle to the direction that it is moving. This
force is doing negative work upon the object.
m.   An individual force does NOT do positive work upon an object if the object is moving at
constant speed.
n.   An object is moving to the right. A force acts leftward upon it. This force is doing
negative work.
o.   A non-conservative force is doing work on an object; it is the only force doing work.
Therefore, the object will either gain or lose mechanical energy.

a. TRUE - Work is a form of energy, and in fact it has units of energy.

b. FALSE - Watt is the standard metric unit of power; Joule is the standard metric unit of
energy.

c. TRUE - A N•m is equal to a Joule.
d. TRUE - A kg•m2/s2 is a mass unit times a speed squared unit, making it a kinetic energy unit
and equivalent to a Joule.

e. FALSE - Work is not dependent on how rapidly the force displaces an object; power is time-
based and calculated by force multiplied by speed.

f. FALSE - Since Superman does not cause a displacement, no work is done; he is merely
holding the car to prevent its descent down the hill.

g. FALSE - The upward force does not cause the horizontal displacement so this is a NON-
example of work.

h. TRUE - There is a component of force in the direction of displacement and so this is an
example of work.

i. TRUE - There is a force and a displacement; the force acts in the opposite direction as the
displacement and so this force does negative work.

j. FALSE - For uniform circular motion, the force acts perpendicular to the direction of the
motion and so the force never does any work upon the object.

k. TRUE - This is clearly work - a force is causing an object to be displaced.

l. FALSE - If a force acts at a 90-degree angle to the direction of motion, then the force does not
do any work at all. Negative work is done when there is a component of force opposite the
direction of motion.

m. FALSE - There are many instances in which an individual force does positive work and yet
the object maintains a constant speed. Consider a force applied to lift an object at constant speed.
The force does positive work. Consider a car moving at constant speed along a level surface. The
force of the road on the tires does positive work while air resistance does and equal amount of
negative work.

n. TRUE - A force which acts in a direction opposite the motion of an object will do negative
work.

o. TRUE - When non-conservative forces do work upon an object, the object will either gain or
lose mechanical energy. Mechanical energy is conserved (neither gained nor lost) only when
conservative forces do work upon objects.

2. Which of the following statements are true about power? Include all that apply.

a. Power is a time-based quantity.
b. Power refers to how fast work is done upon an object.
c. Powerful people or powerful machines are simply people or machines which always do a
lot of work.
d. A force is exerted on an object to move it at a constant speed. The power delivered by
this force is the magnitude of the force multiplied by the speed of the object.
e. The standard metric unit of power is the Watt.
f. If person A and person B do the same job but person B does it faster, then person A does
more work but person B has more power.
g. The Newton•meter is a unit of power.
h. A 60-kg boy runs up a 2.0 meter staircase in 1.5 seconds. His power is approximately 80
Watt.
i. A 300-Newton force is applied to a skier to drag her up a ski hill at a constant speed of
1.5 m/s. The power delivered by the toe rope is 450 Watts.

a. TRUE - Power is a rate quantity and thus time-based.

b. TRUE - This is the definition of power.

c. FALSE - This is not always the case. A machine can do a lot of work but if it fails to do it
rapidly, then it is not necessarily powerful. In fact two machines can do the same task (and
therefore the same work), yet they can have drastically different power ratings.

d. TRUE - An equation for computing work in constant speed situations is P=F•v.

e. TRUE - Watt is the unit of power? Yes!!

f. FALSE - Vice versa. If two people do the same job, then they're doing the same amount of
work. The person who does it fastest generates more power.

g. FALSE - A N•m is a Joule and that is a unit of work (not power). Think force (N) times
distance (m); that's work (J).

h. FALSE - The work would be (m•g)•d or approximately 1200 J. The power is work divided by
time - 1200 J/1.5 s = 800 W.

i. TRUE - Since force and speed are given, use Power = F•v. The calculation yields 450 W.

3. Consider the following physical situations. For each case, determine the angle between the
indicated force (in boldface type) and the displacement ("theta" in the work equation).

A. 0           B. 180            C. 90          D. 30           E. 60
DEGREES         DEGREES          DEGREES        DEGREES         DEGREES

a. A rightward applied force is used to displace a television set to the right. - Answer: A -
0 degrees
b. The force of friction acts upon a rightward-moving car to bring it to a stop. - Answer: B
- 180 degrees
c. A waiter uses an applied force to balance the weight of a tray of plates as he carries the
tray across the room. - Answer: C - 90 degrees
d. The force of air resistance acts upon a vertically-falling skydiver. - Answer: B - 180
degrees
e. The force of friction acts upon a baseball player as he slides into third base. - Answer: B
- 180 degrees
f. An applied force is used by a freshman to lift a World Civilization book to the top shelf
of his locker. - Answer: A - 0 degrees
g. A bucket of water is tied to a string and tension supplies the centripetal force to keep it
moving in a circle at constant speed. - Answer: C - 90 degrees
h. An applied force acting at 30-degrees to the horizontal is used to displace an object to
the right. - Answer: D - 30 degrees
i. A group of football players use an applied force to push a sled across the grass. -
Answer: A - 0 degrees
j. The tension in the elevator cable causes the elevator to rise at a constant speed. -
Answer: A - 0 degrees
k. In a physics lab, an applied force is exerted parallel to a plane inclined at 30-degrees in
order to displace a cart up the incline. - Answer: A - 0 degrees
l. An applied force is exerted upwards and rightwards at an angle of 30-degrees to the
vertical in order to displace an object to the right. - Answer: E - 60 degrees
m. A child rests on the seat of a swing which is supported by the tension in its cables; he
swings from the highest position to its lowest position. - Answer: C - 90 degrees

a.   A - 0 degrees
b.   B - 180 degrees
c.   C - 90 degrees
d.   B - 180 degrees
e.   B - 180 degrees
f.   A - 0 degrees
g.   C - 90 degrees
h.   D - 30 degrees
i.   A - 0 degrees
j.   A - 0 degrees
k.   A - 0 degrees
l.   E - 60 degrees
m.   C - 90 degrees

a. The forward motion is do to the forward pushing; if the force and motion are in the same
direction, then the angle is 0 degrees.

b. Friction opposes motion and as such does negative work; the angle is 180 degrees.

c. The force is vertical and the displacement if horizontal; they make a 90 degree angle.
d. Air resistance opposes motion and as such does negative work; the angle is 180 degrees.

e. Friction opposes motion and as such does negative work; the angle is 180 degrees.

f. The frosh applies an upward force to cause an upward displacement; the angle is 0 degrees.

g. For uniform circular motion, the force is inwards and the displacement at each instant is
tangent to the circle; these two vectors make a 90 degree angle.

h. This is a straightforward question; no tricks here.

i. The forward motion is do to the forward pushing; if the force and motion are in the same
direction, then the angle is 0 degrees.

j. The cable pulls up on the elevator and the elevator is displaced upward; if the force and motion
are in the same direction, then the angle is 0 degrees.

k. The 30-degree angle is the incline angle, not necessarily the angle between F and d. The force
is parallel to the incline and the cart is displaced along the direction of the incline; so the two
vectors are in the same direction and the angle between them is 0 degrees.

l. Compare the wording of this to part h. This one is tricky because the angle between F and d is
60-degrees. If you missed it, reread the question, paying careful attention to the "with the
vertical" part.

m. As the child swings, she traces out a circular arc and as such the tension (centripetal) is
perpendicular to the direction of motion (tangent).

4. Consider the following physical situations. Identify whether the indicated force (in boldface
type) does positive work, negative work or no work.

A. Positive Work        B. Negative Work         C. No Work

Description of Physical Situation                                             +, -, or no Work
a. A cable is attached to a bucket and the force of tension is used to pull
the bucket out of a well.
b. Rusty Nales uses a hammer to exert an applied force upon a stubborn
nail to drive it into the wall.
c. Near the end of the Shockwave ride, a braking system exerts an
applied force upon the coaster car to bring it to a stop.
d. The force of friction acts upon a baseball player as he slides into
third base.
e. A busy spider hangs motionless from a silk thread, supported by the
tension in the thread.
f. In baseball, the catcher exerts an abrupt applied force upon the ball to
stop it in the catcher's mitt.
g. In a physics lab, an applied force is exerted parallel to a plane
inclined at 30-degrees in order to displace a cart up the incline.
h. A pendulum bob swings from its highest position to its lowest
position under the influence of the force of gravity.

Answer: See table below; explanations provided below.

Description of Physical Situation                                             +, -, or no Work
a. A cable is attached to a bucket and the force of tension is used to pull
A. Positive Work
the bucket out of a well.
b. Rusty Nales uses a hammer to exert an applied force upon a stubborn
A. Positive Work
nail to drive it into the wall.
c. Near the end of the Shockwave ride, a braking system exerts an
B. Negative Work
applied force upon the coaster car to bring it to a stop.
d. The force of friction acts upon a baseball player as he slides into
B. Negative Work
third base.
e. A busy spider hangs motionless from a silk thread, supported by the
C. No Work
tension in the thread.
f. In baseball, the catcher exerts an abrupt applied force upon the ball to
B. Negative Work
stop it in the catcher's mitt.
g. In a physics lab, an applied force is exerted parallel to a plane
A. Positive Work
inclined at 30-degrees in order to displace a cart up the incline.
h. A pendulum bob swings from its highest position to its lowest
A. Positive Work
position under the influence of the force of gravity.

a. The force is upwards and the displacement is upwards. When the force and the displacement
act in the same direction, positive work is done.

b. The force is horizontal and the displacement is horizontal. When the force and the
displacement act in the same direction, positive work is done. (It is true that the wall is doing
negative work upon the nail but this statement is about the hammer's force on the nail.)
c. The force is backwards and the displacement is forwards. When the force and the
displacement act in the opposite direction, negative work is done.

d. The force is backwards and the displacement is forwards. When the force and the
displacement act in the opposite direction, negative work is done.

e. If the force does not cause the object to be displaced (the object hangs motionless), then no
work is done.

f. The force is backwards and the displacement is forwards. When the force and the displacement
act in the opposite direction, negative work is done.

g. The force is upwards and parallel to the incline and the displacement is in the same direction
parallel to the incline. When the force and the displacement act in the same direction, positive
work is done.

h. As the bob swings downwards from its highest position, the motion is downwards (and
rightwards); the force is also downwards and as such there is a component of force in the
direction of motion. When the force and the displacement act in the same direction, positive
work is done. (Note that if the bob was swinging upwards from its lowest position to its highest
position, then gravity would be doing negative work.)

5. Which of the following statements are true about conservative(internal) and non-conservative
(external) forces? Include all that apply.

a. A force is regarded as a conservative force if it does work but does not remove
mechanical energy from a system of objects.
b. A force is regarded as a non-conservative force if it does not add mechanical energy to a
system of objects.
c. The force of gravity and elastic (spring) force are both examples of a conservative forces.
d. Applied forces, air resistance, friction forces, and tension are common examples of non-
conservative forces.
e. Physicists envy biologists' ability to instill order on the world of animal species through
their taxonomic system. So physicists have made a habit of identifying forces as
conservative and non-conservative forces in order to instill order on the world of forces.
f. If a non-conservative force acts upon an object, then the object will either gain or lose
mechanical energy.
g. If the only forces which do work upon an object are conservative forces, then the object
will conserve its mechanical energy.
h. If the sum of an object's KE and PE is remaining constant, then non-conservative forces
are NOT doing work.
i. If work is NOT done on an object by a non-conservative force, then the object will
experience a transformation of energy from kinetic to potential energy (or vice versa).
j. An object starts from an elevated position with 50 J of potential energy and begins its fall
towards the ground. If non-conservative forces can be assumed to NOT do work, then at
some point during the fall the object will have 20 J of potential energy and 30 J of kinetic
energy.

Answer: A(sort of) CDGH I(sort of) J

a. TRUE (sort of) - If a force does work, yet does not remove mechanical energy from an object,
then it is definitely a conservative force. The sort of indicates that a force is also considered a
conservative force if it does work and does not add mechanical energy to an object.

b. FALSE - If a force does not add mechanical energy to a system of objects, then it is likely a
conservative force (provided it doesn't remove mechanical energy either). Non-conservative
forces are those which either add or remove energy from a system of objects.

c. TRUE - You must know this!

d. TRUE - These are all non-conservative forces. You can add normal force to the list as well.

e. FALSE - Whether there is envy in a physicist's heart is not for us to tell; the evil found within
one's heart is often vast and mysterious ... . We can however definitively say that a physicist
classifies forces in order to analyze physical situations in accord with the classification. If only
conservative-classified forces do work, then KEi + PEi = KEf + PEf. On the other hand if one or
more non-conservative-classified forces are doing work, then KEi + PEi + Wnc = KEf + PEf.

f. FALSE - Not only must the force act upon the object, it must also be doing work upon the
object. As you sit in your chair, there is a non-conservative force (normal force) acting upon your
body. But since it does not do work (it's being assumed that you are not sitting in one of those
fancy lounge chairs that has more controls than a TV set), your mechanical energy is not
changing.

g. TRUE - This is a big principle. You must know this one!

h. TRUE - Non conservative forces would alter the total mechanical energy; that is, the PE + KE
would not be a constant value.

i. TRUE (sort of) - This statement is true (sort of); when only conservative forces are doing
work, an object has its kinetic energy transformed into potential energy (or vice versa) without
the total amount of the two being altered. It would however be possible that work is not done by
a non-conservative force and there be no transformation of energy at all; i.e., the object remains
at rest. A conservative force must be doing work in order for there to be a transformation of
energy.

j. TRUE - One would notice that the PE would begin to drop from 50 J to 0 J and that the KE
would increase from 0 J to 50 J. And of course there would be a point at which the PE/KE would
be distributed with 20 J to PE and 30 J to KE.
6. Which of the following statements are true about kinetic energy? Include all that apply.

a. Kinetic energy is the form of mechanical energy which depends upon the position of an
object.
b. If an object is at rest, then it does not have any kinetic energy.
c. If an object is on the ground, then it does not have any kinetic energy.
d. The kinetic energy of an object is dependent upon the weight and the speed of an object.
e. Faster moving objects always have a greater kinetic energy.
f. More massive objects always have a greater kinetic energy.
g. Kinetic energy is a scalar quantity.
h. An object has a kinetic energy of 40 J. If its mass were twice as much, then its kinetic
energy would be 80 J.
i. An object has a kinetic energy of 40 J. If its speed were twice as much, then its kinetic
energy would be 80 J.
j. Object A has a mass of 1 kg and a speed of 2 m/s. Object B has a mass of 2 kg and a
speed of 1 m/s. Objects A and B have the same kinetic energy.
k. An object can never have a negative kinetic energy.
l. A falling object always gains kinetic energy as it falls.
m. A 1-kg object is accelerated from rest to a speed of 2.0 m/s. This object gains 4.0 Joules
of kinetic energy.
n. If work is done on an object by a non-conservative force, then the object will either gain
or lose kinetic energy.

a. FALSE - Kinetic energy depends upon the speed of the object; potential energy depends upon
the position of the object.

b. TRUE - Kinetic energy depends upon speed. If there is no speed (the object is at rest), then
there is no kinetic energy.

c. FALSE - If an object is on the ground, then it does not have potential energy (relative to the
ground).

d. FALSE (sort of) - Kinetic energy depends upon mass and speed. Two object's of the same
mass could have different weights if in a different gravitational field; so it is not appropriate to
say that kinetic energy depends upon weight.

e. FALSE - Faster moving objects would have more kinetic energy than other objects of the
same mass. However, another object could have less speed and make up for this lack of speed in
terms of a greater mass.

f. FALSE - More massive objects would have more kinetic energy than other objects with the
same speed. However, another object could have less mass and make up for this lack of mass in
terms of a greater speed.
g. TRUE - Kinetic energy does not have a direction associated with it; it is a scalar quantity.

h. TRUE - Kinetic energy is directly related to the mass of an object.

i. FALSE - Kinetic energy is directly related to the square of the speed of an object. So a
doubling of the speed would result in a quadrupling of the kinetic energy - the new KE would be
160 J.

j. FALSE - When it comes to kinetic energy, speed is doubly important (recall v2). So in this
case, object A would have more kinetic energy. Doing the calculation yields 2 J for object A and
1 J for object B.

k. TRUE - Kinetic energy is determined by the equation 0.5•m•v2. the quantity m is always
positive. And even if v is negative, v2 will always be positive. Therefore, kinetic energy can
never be a negative value.

l. FALSE - If an object is falling at a constant velocity (i.e., the air resistance force equals the
downward force of gravity), then there is not an increase in kinetic energy. It is true however that
free-falling objects always increase their kinetic energy as they fall.

m. FALSE - The kinetic energy increases from 0 J to 2 J (0.5•1•22); that's an increase by 2 J.

n. FALSE - Such an object will definitely gain or lose mechanical energy but not necessarily
kinetic energy.

7. Which of the following statements are true about mechanical energy? Include all that apply.

a. The total amount of mechanical energy of an object is the sum of its potential energy and
the kinetic energy.
b. Heat is a form of mechanical energy.
c. The mechanical energy of an object is always conserved.
d. When non-conservative forces do work, energy is transformed from kinetic to potential
(or vice versa), but the total mechanical energy is conserved.
e. A bowling ball is mounted from a ceiling by way of a strong cable. It is drawn back and
released, allowed to swing as a pendulum. As it swings from its highest position to its
lowest position, the total mechanical energy is mostly conserved.
f. When a friction force does work on an object , the total mechanical energy of that object
is changed.
g. The total mechanical energy of an object remains constant if the only forces doing work
on the object are conservative forces.
h. If an object gains mechanical energy, then one can be certain that a non-conservative
force is doing work.

a. TRUE - This is the definition of mechanical energy.

b. FALSE - Heat or thermal energy is a non-mechanical form of energy. Potential and kinetic
energy are the only forms of mechanical energy.

c. FALSE - The mechanical energy of an object is only conserved if non-conservative forces do
not do work upon the object.

d. FALSE- If a non-conservative force does work upon an object, then the total mechanical
energy of that object is changed. Energy will not be conserved.

e. TRUE - Tension does not do work upon the object and so the total mechanical energy is
conserved. The presence of air resistance (a non-conservative force) does a little work and so one
might notice a very slight change in mechanical energy.

f. TRUE - Friction is a non-conservative force and thus alters the total mechanical energy of an
object.

g. TRUE - This is the conservation of energy principle and one that you need to firmly
understand.

h. TRUE - If there is any change in the total mechanical energy of an object (whether a gain or a
loss), then you know for certain that there is a non-conservative force doing work.

8. Rank these four objects in increasing order of potential energy, beginning with the smallest.

Object A                  Object B                 Object C                 Object D

m = 5.0 kg              m = 10.0 kg                m = 1.0 kg              m = 5.0 kg

v = 4.0 m//s              v = 2.0 m//s              v = 5.0 m/s             v = 2.0 m//s

h = 2.0 m               h = 3.00 m                h = 5.0 m                h = 4.0 m

Answer: C < A < D < B

This is probably best done by performing a calculation of PE and comparing the results:

Object A: PE = (5.0 kg)•(~10 m/s2)•(2.0 m) = 100 J

Object B: PE = (10.0 kg)•(~10 m/s2)•(3.00 m) = 300 J

Object C: PE = (1.0 kg)•(~10 m/s2)•(5.0 m) = 50 J
Object D: PE = (5.0 kg)•(~10 m/s2)•(4.0 m) = 200 J

The order is evident once the calculations are performed

NOTE: The next 15 questions presume that the value of g is 10 m/s/s.

9. A 1200 kg car and a 2400 kg car are lifted to the same height at a constant speed in a auto
service station. Lifting the more massive car requires ____ work.

a. less work                    b. the same                                 c. twice as much
d. four times as much           e. more than 4 times as much

The amount of work done by a force to displace an object is found from the equation

W = F*d*cos(Theta)
The force required to raise the car at constant speed is equivalent to the weight (m*g) of the car.
Since the 2400-kg car weighs 2X as much as the 1200-kg car, it would require twice as much
work to lift it the same distance.

10. An arrow is drawn back so that 50 Joules of potential energy is stored in the stretched bow
and string. When released, the arrow will have a kinetic energy of ____ Joules.

a. 50       b. more than 50         c. less than 50

A drawn arrow has 50 J of stored energy due to the stretch of the bow and string. When
released, this energy is converted into kinetic energy such that the arrow will have 50 J of
kinetic energy upon being fired. Of course, this assumes no energy is lost to air resistance,
friction or any other non-conservative forces and that the arrow is shot horizontally.

11. A child lifts a box up from the floor. The child then carries the box with constant speed to the
other side of the room and puts the box down. How much work does he do on the box while
walking across the floor at constant speed?

a. zero J                    b. more than zero J

For any given situation, the work done by a force can be calculated using the equation

W = F*d*cos(Theta)
where F is the force doing the work, d is the displacement of the object, and Theta is the angle
between the force and the displacement. In this specific situation, the child is applying an upward
force on the box (he is carrying it) and the displacement of the box is horizontal. The angle
between the force (vertical) and the displacement (upward) vectors is 90 degrees. Since the
cosine of 90-degrees is 0, the child does not do any work upon the box.

12. A 1000-kg car is moving at 40 km/hr when the driver slams on the brakes and skids to a stop
(with locked brakes) over a distance of 20 meters. How far will the car skid with locked brakes if
it is traveling at 120 km/hr?

a. 20 m             b. 60 m.           c. 90 m            d. 120 m              e. 180 m

When a car skids to a stop, the work done by friction upon the car is equal to the change in
kinetic energy of the car. Work is directly proportional to the displacement of the car (skidding
distance) and the kinetic energy is directly related to the square of the speed (KE=0.5*m*v^2).
For this reason, the skidding distance is directly proportional to the square of the speed. So if the
speeds is tripled from 40 km/hr to 120 km/hr, then the stopping distance is increased by a factor
of 9 (from 20 m to 9*20 m; or 180 m).

13. A platform diver weighs 500 N. She steps off a diving board that is elevated to a height of 10
meters above the water. The diver will possess ___ Joules of kinetic energy when she hits the
water.

a. 10      b. 500          c. 510        d. 5000          e. more than 5000 .

The use of the work-energy theorem and a simple analysis will yield the solution to this problem.
Initially, there is only PE; finally, there is only KE. Assuming negligible air resistance, the
kinetic energy of the diver upon hitting the water is equal to the potential energy of the diver on
top of the board.

PEi = KEf

m*g*hi = KEf

Substituting 500 N for m*g (500 N is the weight of the diver, not the mass) and 10 m for h will
yield the answer of 5000 J.
14. A ball is projected into the air with 100 J of kinetic energy. The kinetic energy is transformed
into gravitational potential energy on the path towards the peak of its trajectory. When the ball
returns to its original height, its kinetic energy is ____ Joules. Do consider the effects of air
resistance

a. less than 100                        b. 100            c. more than 100
d. not enough information given

During any given motion, if non-conservative forces do work upon the object, then the total
mechanical energy will be changed. If non-conservative forces do negative work (i.e.,
Fnc*d*cos(Theta) is a negative number), then the final TME is less than the initial TME. In this
case, air resistance does negative work to remove energy from the system. Thus, when the ball
returns to its original height, their is less TME than immediately after it was thrown. At this same
starting height, the PE is the same as before. The reduction in TME is made up for by the fact
that the kinetic energy has been reduced; the final KE is less than the initial KE.

15. During a construction project, a 2500 N object is lifted high above the ground. It is released
and falls 10.0 meters and drives a post 0.100 m into the ground. The average impact force on the
object is ____ Newtons.

a. 2500            b. 25000               c. 250,000                 d. 2,500,000

The use of the work-energy theorem and a simple analysis will yield the solution to this problem.
Initially, there is only PE; finally, there is neither PE nor KE; non-conservative work has been
done by an applied force upon the falling object. The work-energy equation can be written as
follows.

PEi + Wnc = 0

PEi = - Wnc

m*g*hi = - F*d*cos(Theta)

Substituting 2500 N for m*g (2500 N is the weight of the driver, not the mass); 10.0 m for h;
0.100 m for the displacement of the falling object as caused by the upward applied force exerted
by the post; and 90 degrees for Theta (the angle between the applied force and the displacement
of the falling object) will yield the answer of 250000 N for F.
16. A 10-Newton object moves to the left at 1 m/s. Its kinetic energy is approximately ____
Joules.

a. 0.5             b. 1            c. 10              d. more than 10

The KE of any object can be computed if the mass (m) and speed (v) are known. Simply use the
equation

KE=0.5*m*v2

In this case, the 10-N object has a mass of approximately 1 kg (use Fgrav = m*g). The speed is 1
m/s. Now plug and chug to yield KE of approximately 0.5 J.

17. Luke Autbeloe stands on the edge of a roof throws a ball downward. It strikes the ground
with 100 J of kinetic energy. Luke now throws another identical ball upward with the same
initial speed, and this too falls to the ground. Neglecting air resistance, the second ball hits the
ground with a kinetic energy of ____ J.

a. less than 100          b. 100       c. 200    d. more than 200            e. none of these

Quite surprisingly to many, each ball would hit the ground with the same speed. In each case, the
PE+KE of the balls immediately after being thrown is the same (they are thrown with the same
speed from the same height). Upon hitting the ground, they must also have the same PE+KE.
Since the PE is zero (on the ground) for each ball, it stands to reason that their KE is also the
same. That's a little physics and a lot of logic - and try not to avoid the logic part by trying to

18. An object at rest may have __________.

a. speed       b. velocity          c. acceleration            d. energy       e. all of these

An object at rest absolutely cannot have speed or velocity or acceleration. However, an object at
rest could have energy if there is energy stored due to its position; for example, there could be
gravitational or elastic potential energy.
19. A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height
(relative to the water) from which the diver dove was approximately ____ meters.

a. 5               b. 10                    c. 50                     d. 100

The kinetic energy of the diver upon striking the water must be equal to the original potential
energy. Thus,

m*g*hi = KEf

(50 kg)*(~10 m/s/s)*h = 5000 J

So, h = ~10 m

20. A job is done slowly, and an identical job is done quickly. Both jobs require the same amount
of ____, but different amounts of ____. Pick the two words which fill in the blanks in their
respective order.

a. energy, work        b. power, work         c. work, energy         d. work, power

e. power, energy       f. force, work         g. power, force         h. none of these

Power refers to the rate at which work is done. Thus, doing two jobs - one slowly and one
quickly - involves doing the same job (i.e., the same work and same force) at different rates or
with different power.

21. Which requires more work: lifting a 50 kg crate a vertical distance of 2 meters or lifting a 25
kg crate a vertical distance of 4 meters?

a. lifting the 50 kg crate                                      b. lifting the 25 kg crate
c. both require the same amount of work

Work involves a force acting upon an object to cause a displacement. The amount of work done
is found by multiplying F*d*cos(Theta). The equation can be used for these two motions to find
the work.

Lifting a 50 kg sack vertically 2 meters           Lifting a 25 kg sack vertically 4 meters
W = (~500 N)*(2 m)*cos(0)                            W = (~250 N)*(4 m)*cos(0)

W = ~1000 N                                          W = ~1000 N

(Note: The weight of a 50-kg object is approx.      (Note: The weight of a 25-kg object is approx.
500 N; it takes 500 N to lift the object up.)       250 N; it takes 250 N to lift the object up.)

Part B: Straightforward Computational Problems

22. Approximate the work required lift a 2.5-kg object to a height of 6.0 meters. PSYW

The work done upon an object is found with the equation

W = F*d*cos(Theta)

In this case, the d=6.0 m; the F=~25 N (it takes approx. 25 N of force to lift a 2.5-kg object), and
the angle between F and d (Theta) is 0 degrees. Substituting these values into the above equation
yields

W = F*d*cos(Theta) = (~25 N)*(6 m)*cos(0) = ~150 J

23. In the It's All Uphill Lab, a force of 20.8 N is applied parallel to the incline to lift a 3.00-kg
loaded cart to a height of 0.450 m along an incline which is 0.636-m long. Determine the work
done upon the cart and the subsequent potential energy change of the cart. PSYW

There are two methods of solving this problem. The first method involves using the equation

W = F*d*cos(Theta)

where F=20.8 N, d=0.636 m, and Theta=0 degrees. Substituting and solving yields

W = F*d*cos(Theta) = (20.8 N)*(0.636 m)*cos(0) = 13.2 J.
The second method is to recognize that the work done in pulling the cart along the incline
changes the potential energy of the cart. The work done equals the potential energy change.
Thus,

W=Delta PE = m*g*(delta h) = (3 kg)*(9.8 m/s/s)*(0.45 m) = 13.2 J

24. A 50-kg hiker ascends a 40-meter high hill at a constant speed of 1.2 m/s. If it takes 400 s to
climb the hill, then determine ... . PSYW

a.   kinetic energy change of the hiker.
b.   the potential energy change of the hiker.
c.   the work done upon the hiker.
d.   the power delivered by the hiker.

a.   (Delta = ) KE = 0 J
b.   (Delta) PE = +20000 J
c.   W = +20000 J
d.   P = 50 Watts

a. The speed of the hiker is constant so there is no change in kinetic energy - 0 J.

b. The potential energy change can be found by subtracting the initial PE (0 J) from the final PE
(m*g*hf). The final potential energy is 20000 J [from (50 kg)*(10 m/s/s)*(40 m)] and the initial
potential energy is 0 J. So Delta PE = +20000 J.

c. The work done upon the hiker can be found using the work-energy theorem. The equation
reduces to

Wnc = PEf

(PEi = 0 J since the hiker starts on the ground; and KEi = KEf since the speed is constant; these
two terms can be dropped from the equation since they are equal). The final potential energy is
20000 J [from (50 kg)*(10 m/s/s)*(40 m)]. So W = +20000 J.

d. The power of the hiker can be found by dividing the work by the time.

P=W/t=(20000 J)/(400 s) = 50 Watts

25. An 800.0-kg car skids to a stop across a horizontal surface over a distance of 45.0 m. The
average force acting upon the car is 7000.0 N. Determine ... . PSYW

a. the work done upon the car.
b. the initial kinetic energy of the car.
c. the acceleration of the car.
d. the initial velocity of the car.

1.   W = -315000 J
2.   KEi = +315000 J
3.   a = -8.75 m/s/s
4.   vi = 28.1 m/s

a. The work done upon the car can be found using the equation

W = F*d*cos(Theta)

where F=7000 N, d=45 m, and Theta=180 degrees (the force is in the opposite direction as the
displacement). Substituting and solving yields -315000 J.

b. The initial kinetic energy can be found using the work-energy theorem. The equation reduces
to

KEi + Wnc = 0

(PEi and PEf = 0 J since the car is on the ground; and KEf = 0 J since the car is finally stopped).
Rearrange the equation and it takes the form KEi = -Wnc . So KEi = +315000 J.

c. The acceleration of the car can be found using Newton's second law of motion: Fnet = m*a

The friction force is the net force (since the up and down forces balance) and the mass is 800 kg.
Substituting and solving yields a = -8.75 m/s/s.

d. The initial velocity of the car can be found using the KE equation: KE = 0.5*m*v2 where
m=800 kg and KEi=315000 J. Substituting and solving for velocity (v) yields v = 28.1 m/s. (A
kinematic equation could be also used to find the initial velocity.)

26. A 60.0-kg skier accelerates down an icy hill from an original height of 500.0 meters. Use the
work-energy theorem to determine the speed at the bottom of the hill if...

a. ... no energy is lost or gained due to friction, air resistance and other non-conservative forces.
PSYW

b. ... 1.40 x 105 J of energy are lost due to external forces. PSYW

Answers: (a) v = 1.00 x 102 m/s; (b) v = 73.0 m/s

a. Use the work energy theorem:
KEi + PEi + Wnc = KEf + PEf

The PEf can be dropped from the equation since the skier finishes on the ground at zero height.
The KEi can also be dropped since the skier starts from rest. The Wnc term is dropped since it is
said that no work is done by non-conservative (external) forces. The equation simplifies to

PEi =KEf

The expressions for KE (0.5*m*v2) and PE (m*g*h) can be substituted into the equation:

m*g*h = 0.5*m*vf2

where m=60 kg, h=500 m, g=10 m/s/s (approx.). Substituting and solving for vf yields 100 m/s.

b. This equation can be solved in a similar manner, except that now the Wext term is -140000J. So
the equation becomes

m*g*h - 140000J = 0.5*m*vf2
Now substituting and solving for vf yields 73.0 m/s.

27. Use the work-energy theorem to determine the force required to stop a 1000.0-kg car moving
at a speed of 20.0 m/s if there is a distance of 45.0 m in which to stop it. PSYW

Answer: F = 4.44*103 N

The work energy theorem can be written as

KEi + PEi + Wnc = KEf + PEf

The PEi and PEf can be dropped from the equation since they are both 0 (the height of the car is 0
m). The KEf can also be dropped for the same reason (the car is finally stopped). The equation
simplifies to

KEi + Wnc = 0

The expressions for KE (0.5*m*v2) and Wnc (F*d*cos[Theta]) can be substituted into the
equation:

0.5*m*vi2 + F*d*cos[Theta] = 0

where m=1000 kg, vi=20 m/s, d=45 m, and Theta = 180 degrees. Substituting and solving for F
yields 4.44*103 N.
Part C: Work-Energy Bar Charts, Analysis, and Conceptual Reasoning

28. Consider the following physical situations. Identify the forces which do work upon the
indicated object (in boldface type) and categorize them as conservative or non-conservative
forces. Then indicate whether the total mechanical energy of the object changes; it it changes,
then indicate whether the change is a positive or negative change. Finally, indicate whether the
potential energy and the kinetic energy changes; if PE or KE changes, then indicate whether the
change is a positive or negative change.

Identity of Forces
Change in    Change in     Change in
TME           PE           KE
Which Do Work:
Description of Physical                          External
Yes           Yes            Yes
Situation                             Internal
(Non-             No             No             No
(Conserv)              (+ or        (+ or -        (+ or -
-)            )              )
Conserv)
a. A force is applied to move a
physics cart from the floor to
the top of an inclined plane at a
constant speed.
b. A physics student scurries
up a flight of stairs at constant
speed.
c. In a moment of unsupervised
phun, a physics student hoists
herself onto a staircase banister
and accelerates down the
banister. Ignore all friction
forces.
d. A ball is dropped from rest
from on the top of a hill and
falls to the ground below.
Ignore air resistance.
e. A ball leaves top of a hill
with a large horizontal velocity.
It falls to the ground below.
Ignore air resistance.
f. A Hot Wheels car is at rest at
an elevated position along an
inclined plane; it is released and
rolls to a position along the
ground. Ignore air resistance.
g. A Hot Wheels car is in
motion at the bottom of a hill
when it hits a computer diskette
box and skids to a stop.
h. A pendulum bob swings
from its highest position to its
lowest position.
i. A physics cart is released
from rest at an elevated position
along an inclined plane; it is
released and rolls to a position
along the incline approximately
5 cm from the bottom.

Identity of Forces
Change in    Change in     Change in
TME           PE           KE
Which Do Work:
Description of Physical                         External
Yes           Yes            Yes
Situation                            Internal
(Non-             No             No             No
(Conserv)              (+ or        (+ or -        (+ or -
-)            )              )
Conserv)
a. A force is applied to move a
physics cart from the floor to
Gravity    Applied     +             +                      No
the top of an inclined plane at a
constant speed.
b. A physics student scurries                   Applied
up a flight of stairs at constant    Gravity      or        +             +                      No
speed.                                          Normal
c. In a moment of unsupervised
phun, a physics student hoists
herself onto a staircase banister
Gravity     None              No      -             +
and accelerates down the
banister. Ignore all friction
forces.
d. A ball is dropped from rest
from on the top of a hill and
Gravity    None              No        -              +
falls to the ground below.
Ignore air resistance.
e. A ball leaves top of a hill
with a large horizontal velocity.
Gravity    None              No        -              +
It falls to the ground below.
Ignore air resistance.
f. A Hot Wheels car is at rest at
an elevated position along an
inclined plane; it is released and Gravity     None              No        -              +
rolls to a position along the
ground. Ignore air resistance.
g. A Hot Wheels car is in
motion at the bottom of a hill
None     Applied     -                      No       -
when it hits a computer diskette
box and skids to a stop.
h. A pendulum bob swings
from its highest position to its    Gravity    None              No        -              +
lowest position.
i. A physics cart is released
from rest at an elevated position
along an inclined plane; it is
Gravity      None              No        -              +
released and rolls to a position
along the incline approximately
5 cm from the bottom.
Answer: See table above

Note that whenever a non-conservative force is doing work (as in a, b, g), the total mechanical
energy is changing.

However, when the only forces which do work are conservative forces, the total mechanical
energy will remaining constant (i.e., be conserved). In such cases, energy will change form from
potential to kinetic energy (or vice versa); yet the total amount of the two forms would not be
changed.

Potential energy can be considered to change if an object changes its height; if the height
decreases, then the potential energy decreases; it the height increases, then the potential energy
increases.

Kinetic energy can be considered to change if an object changes its speed; if the speed decreases,
then the kinetic energy decreases; it the speed increases, then the kinetic energy increases.
29. Several physical situations are described below. For each situation, simplify the work-energy
equation by canceling any zero terms and any energy terms (whether KE or PE) which are
unchanging. Explain each term which gets canceled. The first problem is done as an example.

Simplification of Work-Energy
Description of Physical Situation
Equation
a. A ball starts from rest on top of
a tall pillar and falls to the ground
below. Assume the effect of air
resistance is negligible.

b. A ball starts from
rest at an elevated
position along an
inclined plane and
rolls to the ground
below. Assume that
the effect of friction
and air resistance is
negligible.

c. A ball starts from
rest at an elevated
position along an
inclined plane and
rolls to the ground
below. Consider the
effect of friction and
air resistance.
d. A track is constructed by
stretching a grooved and
pliable material between two
lab poles. A metal ball starts
from rest at point A and rolls
to point B. Friction and air
resistance have an effect upon the ball's motion.

e. A pendulum bob is
mounted on top of a
lab pole and drawn
back to a string which
it tied between two
other poles. The pendulum bob is released from rest.
Upon reaching the vertical, the string hits a barrier and
a new pivot point is established as the bob continues in
motion along an upward trajectory. Ignore the effect of
air resistance.
f. A Hot Wheels car starts
from rest on top of an
inclined plane and rolls down
the incline through a loop
and along a horizontal
surface. Friction and air
resistance have a significant effect on the car.

g. An unattended hot dog wagon
starts from rest and rolls down a
hill and up a second hill. Ignore
the effect of friction and air
resistance.
h. A roller coaster car is
already in motion on the
top of the first drop and
rolls along the track over
a couple of hills. Ignore
the effect of friction and air resistance.

i. A cross-
country skier
is in motion
on top of a
small hill. He
skis down the
hill into the valley and up a second smaller hill. He
uses his poles to propel himself during the entire
motion. Ignore the effect of friction and air resistance.

Simplification of Work-Energy
Description of Physical Situation
Equation
a. A ball starts from rest on top of                   PEi = KEf
a tall pillar and falls to the ground
below. Assume the effect of air                  (Since initially at rest,
resistance is negligible.                         KEi = 0 and cancels.
Since the final height is
0, PEf = 0 and cancels.
Since non-conservative
forces are not doing
work, Wnc = 0)

b. A ball starts from                    PEi = KEf
rest at an elevated
position along an         (Initially the ball is at rest, KEi = 0 and
inclined plane and        cancels. Since the final height is 0, PEf
rolls to the ground       = 0 and cancels. Since non-conservative
below. Assume that          forces are not doing work, Wnc = 0)
the effect of friction
and air resistance is
negligible.
c. A ball starts from                PEi + Wnc = KEf
rest at an elevated
position along an        (Initially the ball is at rest, KEi = 0 and
inclined plane and       cancels. Since the final height is 0, PEf
rolls to the ground         = 0 and cancels. Friction is a non-
below. Consider the      conservative forces and it does work so
effect of friction and             Wnc does not cancel.)
air resistance.

d. A track is constructed by                  PEi + Wnc = PEf
stretching a grooved and
pliable material between two      (The ball is at rest in both its initial and
lab poles. A metal ball starts   final states, so both KEi = 0 and KEf = 0
from rest at point A and rolls    and cancels. Friction and air resistance
to point B. Friction and air     are non-conservative forces and do work
resistance have an effect upon the ball's motion.                   so Wnc does not cancel.)

e. A pendulum bob is                  PEi = KEf + PEf
mounted on top of a
lab pole and drawn       (Initially the ball is at rest, KEi = 0 and
back to a string which      cancels. At the point shown in the
it tied between two       diagram as the final state, the ball has
other poles. The pendulum bob is released from rest.      both height and would still be moving;
Upon reaching the vertical, the string hits a barrier and    so neither final energy term would
a new pivot point is established as the bob continues in cancel. The only non-conservative force
motion along an upward trajectory. Ignore the effect of     present - tension - does not do work
air resistance.                                           since it is directed at 90-degrees to the
direction of motion; so Wnc cancels.)
f. A Hot Wheels car starts                PEi + Wnc = KEf
from rest on top of an
inclined plane and rolls down (Initially the car is at rest, KEi = 0 and
the incline through a loop    cancels. Since the final height is 0, PEf
and along a horizontal          = 0 and cancels. Friction is a non-
surface. Friction and air     conservative forces and it does work so
resistance have a significant effect on the car.                   Wnc does not cancel.)

g. An unattended hot dog wagon                    PEi = KEf + PEf
starts from rest and rolls down a
hill and up a second hill. Ignore       (Initially the wagon is at rest, KEi = 0
the effect of friction and air        and cancels. The wagon has height in its
resistance.                              initial state, so the PEi term does not
cancel. At the point shown in the
diagram as the final state, the wagon has
both height and would still be moving;
so neither final energy term would
cancel. There is no friction nor air
resistance and the normal force does not
do work since it is perpendicular to the
displacement; so Wnc cancels.)
h. A roller coaster car is                PEi = KEf + PEf
already in motion on the
top of the first drop and   (Initially the car is at rest, KEi = 0 and
rolls along the track over cancels. The car has height in its initial
a couple of hills. Ignore state, so the PEi term does not cancel. At
the effect of friction and air resistance.                   the point shown in the diagram as the
final state, the car has both height and
would still be moving; so neither final
energy term would cancel. There is no
friction nor air resistance and the normal
force does not do work since it is
perpendicular to the displacement; so
Wnc cancels.)
i. A cross-         KEi + PEi + Wnc = KEf + PEf
country skier
is in motion      (Initially, the skier has height and
on top of a      motion and so neither initial energy
small hill. He term will cancel. The skier is using his
skis down the poles to propel himself so there is a non-
hill into the valley and up a second smaller hill. He    conservative force doing work; the Wnc
uses his poles to propel himself during the entire       term does not cancel. In the final state,
motion. Ignore the effect of friction and air resistance.      the skier is moving and has height
(presuming that the zero level is the
valley) and so neither final energy term
will cancel.

Part D: Complex Analysis and Problem-Solving

30. A 21.3-kg child positions himself on an inner-tube which is suspended by a 7.28-m long rope
attached to a strong tree limb. The child and tube are drawn back until it makes a 17.4-degree
angle with the vertical. The child is released and allowed to swing to and from. Assuming
negligible friction, determine the child's speed at his lowest point in the trajectory. PSYW

This is an example of energy transformation from potential energy at the highest point (the point
of release) to kinetic energy at the lowest position. Since gravity is the only force doing work
(tension acts perpendicular to the displacement so it does not do work), the total mechanical
energy is conserved. So the energy conservation equation will be used.

KEi + PEi = KEf + PEf

Since the child starts from rest, the KEi term can be canceled. And if we assign the lowest
position as the zero-level, then the PEf term can be canceled from the equation. The equation can
be rewritten as

PEi = KEf

By substituting in expressions for kinetic and potential energy, the equation becomes

m • g • hi = 0.5 • m • vf2

The mass cancels from each side of the equation, and algebra can be used to manipulate the
equation to solve for vf.

vf = SQRT( 2 • g • hi )

The crux of the problem becomes determining the
initial height of the child. The length of the cables and
the angle to which it is drawn back will be essential
information in determining the initial height. The
diagram at the right depicts the physical situation. The
length of the cable is the hypotenuse of a triangle; the
length of the side adjacent to the 17.4-degree angle
can be determined using the cosine function. This
adjacent side (a) is shorter than the length of the cable by an amount equal to the initial height
(hi). The work is shown below:

cos(17.4 deg) = a/L

a = L • cos(17.4 deg) = 6.95 m

hi = L - a = 7.28 m - 6.95 mm = 0.333 m

Once hi is found, it can be substituted into the above equation to solve for the final velocity:

vf = SQRT ( 2 • 9.8 m/s/s • 0.33 m) = SQRT(6.53 m/s) = 2.56 m/s

31. A baseball player catches a 163-gram baseball which is moving horizontally at a speed of
39.8 m/s. Determine the force which she must apply to the baseball if her mitt recoils a
horizontal distance of 25.1 cm. PSYW

This is an example of work being done by a non-conservative force (the applied force of the
mitt) upon a baseball in order to change its kinetic energy. So

Wnc = Change in KE

The change in kinetic energy can be computed by subtracting the initial value (0.5 • m • vi2) from
the final value (0 J) .

Change in KE = KEf - KEi = 0 J - 0.5 • (0.163 kg) • (39.8 m/s)2 = -129 J

The force can be determined by setting this value equal to the work and using the expression for
work into the equation:

Wnc = -129 J

F • d • cos(theta) = -129 J

F • (0.251 m) • cos(180 deg) = -129 J

F = 514 N

32. A 62.9-kg downhill skier is moving with a speed of 12.9 m/s as he starts his descent from a
level plateau at 123-m height to the ground below. The slope has an angle of 14.1 degrees and a
coefficient of friction of 0.121. The skier coasts the entire descent without using his poles; upon
reaching the bottom he continues to coast to a stop; the coefficient of friction along the level
surface is 0.623. How far will he coast along the level area at the bottom of the slope? PSYW

During the entire descent down the hill, gravity is doing work on the skier and friction is doing
negative work on the skier. Friction is a non-conservative force and will alter the total
mechanical energy of the skier. The equation to be used is

KEi + PEi + Wnc = KEf + PEf

If we designate the level area at the bottom of the slope as the zero level of potential energy, then
PEf is 0 J. Since the skier eventually stops (due to the effect of friction along the level area), the
KEf is 0 J. So the above equation becomes

KEi + PEi + Wnc = 0

The Wnc term has two parts; there is friction doing along the inclined plane and friction doing
work along the level surface. Since these two sections of the motion have different normal forces
and friction coefficients (and therefore friction forces), they will have to be treated separately.
The graphic below depicts the free-body diagrams and the means by which the friction force can
be determined.

By substituting values of mu and mass and g and theta into the above equations, one finds that
the friction values are

On Incline                       On Level Surface

Ffrict = 72.3 N                      Ffrict = 384 N

These forces act upon the skier over different distances. In the case of the inclined plane, the
distance (d) can be computed from the given incline angle and the initial height. The relationship
is depicted in the diagram below. The sine function is used to relate the angle to the initial height
and the distance along the incline. In the case of the level surface, the distance is the unknown
quantity (x) which this problem calls for.
The distance d along the incline is

d = hi / sin(theta) = 123 m / sin(14.1 deg) = 505 m

Now substitutions can be made into the work-energy equation and algebraic manipulation can be
performed to solve for x:

KEi + PEi + Wnc = 0

KEi + PEi + Wincline + Wlevel = 0

0.5•(62.9 kg)•(12.9 m/s)2 + (62.9 kg)•(9.8 m/s)•(123 m) + (72.3 N)•(505 m)•cos(180 deg) + (384
N)•(x)•cos(180 deg) = 0 J

5233 J + 75820 J - 36512 J - 384 x = 0 J

44541 J = 384 x

116 m = x

33. A 221-gram ball is thrown at an angle of 17.9 degrees and a speed of 36.7 m/s from the top
of a 39.8-m high cliff. Determine the impact speed of the ball when it strikes the ground. Assume
negligible air resistance. PSYW

Here is a situation in which the only force doing work
-gravity - is a conservative force; so the total
mechanical energy of the system is conserved. The
conservation of energy equation can be used.

KEi + PEi = KEf + PEf

If we assign the ground below the cliff to be the zero-
level of potential energy, then the PEf term can be
canceled from the equation.

KEi + PEi = KEf
Expressions for kinetic and potential energy can be substituted into the above equation to yield.

0.5•m•vi2 + m•g•hi = 0.5•m•vf2

By dividing each term of the equation by m, the mass can be canceled.

0.5 • vi2 + g • hi = 0.5 • vf2

Now known values can be substituted into this equation and the final velocity can be determined.
The work is shown here:

0.5•(36.7 m/s)2 + (9.8 m/s/s)•(39.8 m) = 0.5 • vf2

673.4 m2/s2 + 390.0 m2/s2 = 0.5 • vf2

1063 m2/s2 = 0.5 • vf2

2127 m2/s2 = vf2

SQRT(2127 m2/s2) = vf

vf = 46.1 m/s

34. Claire deAisles has just completed her shopping at the Jewel food store. She accidentally
bumps her 42.5-kg cart, setting it in motion from rest down a hill inclined at 14.9 degrees. Upon
descending a distance of 9.27 meters along the inclined plane, the cart hits a tree stump (which
was placed in the parking lot for the sole purpose of this problem). A 0.295-kg can of tomato
soup is immediately hurled from the moving cart and heads towards Will N. Tasue's brand new
Lexus. Upon striking the Lexus, the tomato soup can creates a dent with a depth of 3.16 cm.
Noah Formula, who is watching the entire incident and fixing to do some physics, attempts to
calculate the average force which the Lexus applies to the soup can. Assume negligible air
resistance and friction forces and help Noah out. PSYW

Here's an example of a physical situation in which gravity
first acts to do work and convert the potential energy of a
soup can (and the grocery cart) into kinetic energy; then a
non-conservative force - the force of the Lexus applied to
the soup can - serves to change the kinetic energy of the
soup can. The problem is best analyzed if the cart itself is ignored and the focus becomes the
soup can. The soup can begins with potential energy on top of the hill and finishes with no
mechanical energy. The situation is depicted at the right.
The relevant equation becomes the work-energy equation:

KEi + PEi + Wnc = KEf + PEf

The final potential and kinetic energy can be canceled and the initial kinetic energy can be
canceled (the can starts from rest).

PEi + Wnc = 0 J

The expressions for potential energy and work can be substituted into the above equation to
derive:

m•g•hi + F•d•cos(theta) = 0 J

The initial height of the soup can is related to the angle of incline and the length along the incline
according to the equation

sin(theta) = hi/L

The values of theta (14.9 degrees) and L (9.27 m) can be substituted into this equation and hi is
found to be 2.38 m.

This hi value can be substituted into the work-energy equation along with the values of m (0.295
kg) and d (0.0316 m). The force can then be calculated. The work is shown here.

(0.295 kg)•(9.8 m/s/s)•(2.38 m) + F•(0.0316 m)•cos(180 deg) = 0 J

6.89 J - (0.0316m ) • F = 0 J

6.89 J = (0.0316m ) • F

F = 218 N

(Note there is an assumption in this problem that the collision with the Lexus chassis brings the
soup can to a final rest position; This is likely not the case since a soup can typically does not
become lodged inside the Lexus, but rather rebounds with a post-impact velocity. Yet without
this assumption or further information, the problem cannot be solved.)

35. Mia Kneezhirt jumps from a second story dorm room (h = 7.91 m) to the ground below.
Upon contact with the ground, she allows her 62.4-kg body to come to an abrupt stop as her
center of gravity is displaced downwards a distance of 89.2 cm. Calculate the average upward
force exerted by the ground upon Mia's fragile body. PSYW (Use Conservation of Energy
Equation)

This is another instance of a non-conservative force (the force of the ground upon Mia's body)
doing work in order to alter the total mechanical energy. The relevant equation is

KEi + PEi + Wnc = KEf + PEf

Since Mia starts from rest, the KEi term is 0 J and can be canceled. If we designate the ground to
be the zero level, then the PEf term is 0 J and can be canceled as well. Since Mia is finally
stopped by the upward applied force of the ground, the KEf term is 0 J and can be canceled. The
original equation then simplifies to

PEi + Wnc = 0 J

Expressions for potential energy and work can be substituted into this equation to change its
form to

m•g•hi + F•d•cos(theta) = 0 J

All quantities are known except for F, so values can be substituted into the equation and algebra
can be used to solve for F. The work is shown here.

(62.4 kg) • (9.8 m/s/s) • (7.91 m) + F • (0.892 m) • cos(180 deg) = 0

4837 J - (0.892 m) • F = 0

4837 J = (0.892 m) • F

F = 5423 N

(Note that a critical assumption is made to make this problem solve-able. It is assumed that the
Mia's center of mass starts a distance of 7.91 m above the position at which Mia's center of mass
ultimately comes to rest. For most situations, this is usually a very safe assumption. In this case,
the bending of Mia's knees would actually lower the center of gravity lower than the height that
it would be at when she is just standing on the ground (in this case, 0.892 cm lower). Either this
additional distance must be included in the PE term or it is assumed that Mia's landing involves
somewhat of a spring action whereby she bends her knees to lower her body and then straightens
up to an upright position - similar to the landing of an Olympic gymnast. The assumption is
made in this problem that Mia utilizes the ground's force to not only stop her motion but to also
elevate her back up to an upright position.)

36. In a physics lab, a 0.500-kg cart moving at 36.4 cm/s collides inelastically with a second cart
which is initially at rest. The two carts move together with a speed of 21.8 cm/s after the
collision. Determine the mass of the second cart.
Answer: ~ 0.335 kg

This problem involves a perfectly inelastic collision between two carts. Thus, the post-collision
velocity of the two carts are identical. For communication sake, the carts will be referred to as
Cart A and Cart B. The given information is:

mA = 0.500 kg; vA-before = 36.4 cm/s; vB-before = 0 cm/s; vA-after = 21.8 cm/s; vB-after = 21.8 cm/s

The unknown to be solved for in this problem is the mass of Cart B (mB).

The solution begins by setting the writing expressions for the total momentum of the system
before and after the collision.

Before Collision: ptotal-before = (0.500 kg)•(36.4 cm/s) + (mB)•(0 cm/s)

After Collision: ptotal-after = (0.500 kg)•(21.8 cm/s) + (mB)•(21.8 cm/s)

Assuming momentum conservation, these expressions are set equal to each other and then
algebraically manipulated to solve for the unknown (mB).

(0.500 kg)•(36.4 cm/s) + (mB)•(0 cm/s) = (0.500 kg)•(21.8 cm/s) + (mB)•(21.8 cm/s)

(0.500 kg)•(36.4 cm/s) = (0.500 kg)•(21.8 cm/s) + (mB)•(21.8 cm/s)

(0.500 kg)•(36.4 cm/s) - (0.500 kg)•(21.8 cm/s) = (mB)•(21.8 cm/s)

(7.30 kg•cm/s) = (mB)•(21.8 cm/s)

0.33486 kg = mB

mB = ~0.335 kg

37. A classic physics demonstration involves firing a bullet into a block of wood suspended by
strings from the ceiling. The height to which the wood rises below its lowest position is
mathematically related to the pre-collision speed of the bullet. If a 10.3-gram bullet is fired into
the center of a 1.5-kg block of wood and it rises upward a distance of 25 cm, then what was the
pre-collision speed of the bullet?

Answer: 3.2 x 102 m/s

Here is another instance in which momentum principles must be combined with content learned
in other units in order to complete an analysis of a physical situation. The collision involves the
inelastic collision between a block of wood and bullet. The bulled lodges into the wood and the
two objects move with identical velocity after the collision. The kinetic energy of the wood and
bullet is then converted to potential energy as the combination of two objects rises to a final
resting position.

Energy conservation can be used to determine the velocity of the wood-bullet combination
immediately after the collision. The kinetic energy of the wood-bullet combination is set equal to
the final potential energy of the wood-bullet combination and the equation is manipulated to
solve for the post-collision velocity of the wood-bullet combination. The work is shown here:

0.5 • (mwood + mbullet) • vcombination-after2 = (mwood + mbullet) • (9.8 m/s2) • (0.25 m)

0.5 • vcombination-after2 = (9.8 m/s2) • (0.25 m)

vcombination-after2 = 2 •(9.8 m/s2) • (0.25 m)

vcombination-after2 = 4.9 m2/s2

vcombination-after = 2.2135 m/s

Now momentum conservation can be used to determine the pre-collision velocity of the bullet
(vbullet-before). The known information is:

mwood = 1.5 kg; mbullet = 10.3 g = 0.0103 kg; vwood-after = 2.2135 m/s; vbullet-after = 2.2135 m/s

Expressions for the total system momentum can be written for the before- and after-collision
situations.

Before Collision: ptotal-before = (1.5 kg)•(0 m/s) + (0.0103 kg)•(vbullet-before)

After Collision: ptotal-after = (1.5 kg)•(2.2135 m/s) + (0.0103 kg)•(2.2135 m/s)

Assuming momentum conservation, these expressions are set equal to each other and then
algebraically manipulated to solve for the unknown (mB).

(1.5 kg)•(0 m/s) + (0.0103 kg)•(vbullet-before) = (1.5 kg)•(2.2135 m/s) + (0.0103 kg)•(2.2135 m/s)

(0.0103 kg)•(vbullet-before) = (1.5 kg)•(2.2135 m/s) + (0.0103 kg)•(2.2135 m/s)

(0.0103 kg)•(vbullet-before) = 3.3430 kg•m/s

vbullet-before = (3.3430 kg•m/s) / (0.0103 kg)

vbullet-before = 324.5678 m/s

vbullet-before = ~3.2 x 102 m/s
38. On Okanagan Lake, Mr. MacKenzie (108 kg) and his brother (92 kg) are cruising around in
Ogopogo shaped boats (160 kg). Mr. MacKenzie’s gets pedaling at 1.0 m/s and collides head on
with his brother who is moving in the opposite direction at 1.2 m/s. After the collision, his
brother bounces backwards at 0.30 m/s. Assuming an isolated system, determine

a. ... Mr. MacKenzie's post-collision speed.

b. ... the percentage of original kinetic energy which is lost as the result of the collision

Answer: (a) v = ~0.41 m/s

(b) % KE Loss = ~89%

(a) Expressions for the total momentum of the system before and after the collision can be
written. For the before-collision expression, his brother is assigned a positive velocity value and
Mr. MacKenzie is assigned a negative velocity value (since he is moving in the opposite
direction). Furthermore, the mass of the bumper car must be figured into the total mass of the
individually moving objects.

ptotal-before = pbro-before + pMac-before

ptotal-before = mbro • vbro-before + mMac • vMac-before

ptotal-before = (252 kg) • (1.2 m/s) + (268 kg) • (-1.0 m/s)

For the before-collision expression, bro is assigned a negative velocity value (since he has
bounced backwards in the opposite direction of his original motion. Mac is assigned a velocity of
v since his velocity is not known.

ptotal-after = pBro-after + pMac-after

ptotal-after = mBro • vBro-after + mMac • vMac-after

ptotal-after = (252 kg) • (-0.3 m/s) + (268 kg) • (vMac-after)

ptotal-after = (252 kg) • (-0.3 m/s) + (268 kg) • v

Since the system is assumed to be isolated, the before-collision momentum expression can be set
equal to the after-collision momentum expression. The equation can then be algebraically
manipulated to solve for the post-collision velocity of Mr. Mac.

(252 kg) • (1.2 m/s) + (268 kg) • (-1.0 m/s)) = (252 kg) • (-0.3 m/s) + (268 kg) • v
302.4 kg•m/s - 268 kg•m/s = -75.6 + (268 kg) • v

34.4 kg•m/s = - 75.6 kg•m/s + (268 kg) • v

v = 0.4104 m/s = ~0.41 m/s

(b) This collision is neither perfectly elastic (since the collision force is a contact force) nor
perfectly inelastic (since the objects do not stick together). It is a partially elastic/inelastic
collision. Since the collision is not perfectly elastic, there is a loss of total system kinetic energy
during the collision. The before-collision and after-collision kinetic energy values can be
calculated and the percentage of total KE lost can be determined.

The before-collision KE is based on before-collision speeds:

KEsystem-before = KEBro-before + KEMac-before

KEsystem-before = 0.5 • mBro • vBro-before2 + 0.5 • mMac • vMac-before2

KEsystem-before = 0.5 • (252 kg) • (1.2 m/s)2 + 0.5 • (268 kg) • (1.0 m/s)2

KEsystem-before = 181.44 J + 134 J

KEsystem-before = 315.44 J

The after-collision KE is based on after -collision speeds:

KEsystem-after = KEBro-after + KEMac-after

KEsystem-after = 0.5 • mBro • vBro-after + 0.5 • mMac• vMac-after

KEsystem-after = 0.5 • (252 kg) • (0.3 m/s)2 + 0.5 • (268 kg) • (0.4104 m/s)2

KEsystem-after = 11.34 J + 22.5693 J

KEsystem-after = 33.9093 J

The system kinetic energy is changed from 315.44 J to 33.9093 J during the collision. The total
KE lost is 281.5307 J. This value can be used to determine the percent of the original KE which
is lost in the collision.
% KE Loss = (281.5307 J) / (315.44 J) • 100%

% KE Loss = 89.25% = ~89%

39. Two carbon (ultra hard and virtually undeformable) cars collide elastically on a level, low-
friction track. Car A has a mass of 1675 kg and is moving east at 15.0 m/s. Car B has a mass of
1350 kg and is moving West at 12.0 m/s. Determine the post-collision velocities of the two cars.

Answer: vA-after = -9.10 cm/s; vB-after = 17.9 cm/s

This is a perfectly elastic collision in which both momentum and kinetic energy are conserved.
Two equations will be developed using the momentum conservation and kinetic energy
conservation principles. One equation will be used to develop an expression for vA in terms of
vB. This expression will then be substituted into the second equation in order to solve for vB. The
original vA expression can then be used to determine the vA value. The solution is shown below.

The momentum conservation equation can be written as

mA • vA-before + mB • vB-before = mA • vA-after + mB • vB-after

(1675 kg) • (+15.0 m/s) + (1350 kg) • (-12.0 cm/s) = (1675 kg) • vA-after + (1350 kg) • vB-after

8925 kg • m/s = (1675 kg) • vA-after + (1350 kg) • vB-after

For elastic collisions, total system kinetic energy is conserved. The kinetic energy conservation
equation is written as

0.5 • m • vA-before2 + 0.5 • m • vB-before2 = 0.5 • m • vA-after2 + 0.5 • m • vB-after2

As shown in the book, this equation can be simplified to the form of

vA-before + vA-after = vB-before + vB-after

15.0 m/s + vA-after = -12.0 m/s + vB-after

vA-after = vB-after – 27.0 m/s

Now the problem has been reduced to two equations and two unknowns. Such a problem can be
solved in numerous ways. Note that equation 2 represents an expression for vA-after in terms of vB-
after. This expression for vA-after can then be substituted into Equation 1. The value of vB-after can
then be determined. This work is shown below. (To simplify the mathematics, the units will be
dropped from the numerical values stated in the solution. When vB-after is solved for, its units will
be in cm/s - the same units used for velocity in the above portion of the solution.)

8925 = (1675) • (vB-after – 27.0) + (1350) • vB-after
8925 = 1675 • vB-after - 45225 + 1350 • vB-after

54150= 3025 • vB-after

vB-after = 17.9008 m/s

Now that the value of vB-after has been determined, it can be substituted into the original
expression for vA-after (Equation 2) in order to determine the numerical value of vA-after. This is
shown below.

vA-after = vB-after – 27.0 m/s

vA-after = 17.9008 m/s – 27.0 m/s

vA-after = -9.0991 = -9.10 m/s
MOMENTUM REVIEW
Part A: Multiple Multiple Choice

1. Which of the following statements are true about momentum?

a. Momentum is a vector quantity.
b. The standard unit on momentum is the Joule.
c. An object with mass will have momentum.
d. An object which is moving at a constant speed has momentum.
e. An object can be traveling eastward and slowing down; its momentum is westward.
f. Momentum is a conserved quantity; the momentum of an object is never changed.
g. The momentum of an object varies directly with the speed of the object.
h. Two objects of different mass are moving at the same speed; the more massive object
will have the greatest momentum.
i. A less massive object can never have more momentum than a more massive object.
j. Two identical objects are moving in opposite directions at the same speed. The forward
moving object will have the greatest momentum.
k. An object with a changing speed will have a changing momentum.

2. Which of the following are true about the relationship between momentum and energy?

a. Momentum is a form of energy.
b. If an object has momentum, then it must also have mechanical energy.
c. If an object does not have momentum, then it definitely does not have mechanical energy
either.
d. Object A has more momentum than object B. Therefore, object A will also have more
kinetic energy.
e. Two objects of varying mass have the same momentum. The least massive of the two
objects will have the greatest kinetic energy.

3. Which of the following statements are true about impulse?

a. Impulse is a force.
b. Impulse is a vector quantity.
c. An object which is traveling east would experience a westward directed impulse in a
collision.
d. Objects involved in collisions encounter impulses.
e. The Newton is the unit for impulse.
f. The kg•m/s is equivalent to the units on impulse.
g. An object which experiences a net impulse will definitely experience a momentum
change.
h. In a collision, the net impulse experienced by an object is equal to its momentum change.
i. A force of 100 N acting for 0.1 seconds would provide an equivalent impulse as a force
of 5 N acting for 2.0 seconds.

4. Which of the following statements are true about collisions?

a. Two colliding objects will exert equal forces upon each other, even if their mass is
significantly different.
b. During a collision, an object always encounters an impulse and a change in
momentum.
c. During a collision, the impulse which an object experiences is equal to its velocity
change.
d. The velocity change of two respective objects involved in a collision will always be
equal.
e. While individual objects may change their velocity during a collision, the overall or
total velocity of the colliding objects is conserved.
f. In a collision, the two colliding objects could have different acceleration values.
g. In a collision between two objects of identical mass, the acceleration values could be
different.
h. Total momentum is always conserved between any two objects involved in a
collision.
i. When a moving object collides with a stationary object of identical mass, the
stationary object encounters the greater collision force.
j. When a moving object collides with a stationary object of identical mass, the
stationary object encounters the greater momentum change.
k. A moving object collides with a stationary object; the stationary object has
significantly less mass. The stationary object encounters the greater collision force.
l. A moving object collides with a stationary object; the stationary object has
significantly less mass. The stationary object encounters the greater momentum
change.

5. Which of the following statements are true about elastic and inelastic collisions?

a. Perfectly elastic and perfectly inelastic collisions are the two opposite extremes along a
continuum; where a particular collision lies along the continuum is dependent upon the
amount kinetic energy which is conserved by the two objects.
b. Most collisions tend to be partially to completely elastic.
c. Momentum is conserved in an elastic collision but not in an inelastic collision.
d. The kinetic energy of an object remains constant during an elastic collision.
e. Elastic collisions occur when the collision force is a non-contact force.
f. Most collisions are not inelastic because the collision forces cause energy of motion to be
transformed into sound, light and thermal energy (to name a few).
g. A ball is dropped from rest and collides with the ground. The higher that the ball rises
upon collision with the ground, the more elastic that the collision is.
h. A moving air track glider collides with a second stationary glider of identical mass. The
first glider loses all of its kinetic energy during the collision as the second glider is set in
motion with the same original speed as the first glider. Since the first glider lost all of its
kinetic energy, this is a perfectly inelastic collision.
i. The collision between a tennis ball and a tennis racket tends to be more elastic in nature
than a collision between a halfback and linebacker in football.

Part B: Multiple Choice

6. Which of the following objects have momentum? Include all that apply.

a. an electron orbiting the nucleus of an atom.

b. a UPS truck stopped in front of the school building.

c. a Yugo (a compact car) moving with a constant speed.

d. a small flea moving with constant speed across Fido's back.

e. Glenbrook South High School.

Answer: A, C, and D

7. A truck driving along a highway road has a large quantity of momentum. If it moves at the
same speed but has twice as much mass, its momentum is ________________.

a. zero          b. quadrupled                   c. doubled            d. unchanged

8. TRUE or FALSE:

A ball is dropped from the same height upon various flat surfaces. For the same
collision time, impulses are smaller when the most bouncing take place.

a. True                                           b. False

9. Consider a karate expert. During a talent show, she executes a swift blow to a cement block
and breaks it with her bare hand. During the collision between her hand and the block, the ___.

a.   time of impact on both the block and the expert's hand is the same
b.   force on both the block and the expert's hand have the same magnitude
c.   impulse on both the block and the expert's hand have the same magnitude
d.   all of the above.
e.   none of the above.

10. It is NOT possible for a rocket to accelerate in outer space because ____. List all that apply.

a.   there is no air in space
b.   there is no friction in space
c.   there is no gravity in outer space
d.   ... nonsense! Rockets do accelerate in outer space.

11. In order to catch a ball, a baseball player naturally moves his or her hand backward in the
direction of the ball's motion. This habit causes the force of impact on the players hand to be
reduced in size principally because ___.

a.   the resulting impact velocity is lessened
b.   the momentum change is decreased
c.   the time of impact is increased
d.   the time of impact is decreased
e.   none of these

12. Suppose that Paul D. Trigger fires a bullet from a gun. The speed of the bullet leaving the
muzzle will be the same as the speed of the recoiling gun ____.

a.   because momentum is conserved
b.   because velocity is conserved
c.   because both velocity and momentum are conserved
d.   only if the mass of the bullet equals the mass of the gun
e.   none of these

13. Suppose that you're driving down the highway and a moth crashes into the windshield of
your car. Which undergoes the greater change is momentum?

a. the moth                 b. your car                   c. both the same

14. Suppose that you're driving down the highway and a moth crashes into the windshield of
your car. Which undergoes the greater force?

a. the moth                 b. your car                   c. both the same

15. Suppose that you're driving down the highway and a moth crashes into the windshield of
your car. Which undergoes the greater impulse?

a. the moth                 b. your car                   c. both the same

16. Suppose that you're driving down the highway and a moth crashes into the windshield of
your car. Which undergoes the greater acceleration?

a. the moth                 b. your car                   c. both the same

17. Three boxes, X, Y, and Z, are at rest on a
table as shown in the diagram at the right. The
weight of each box is indicated in the diagram.
The net or unbalanced force acting on box Y is
_____.
a. 4 N down            b. 5 N down            c. 5 N up          d. 10 N up       e. zero

18. In a physics experiment, two equal-mass Pasco carts roll towards each other on a level, low-
friction track. One cart rolls rightward at 2 m/s and the other cart rolls leftward at 1 m/s. After
the carts collide, they couple (attach together) and roll together with a speed of _____________.
Ignore resistive forces.

a. 0.5 m/s        b. 0.33 m/s          c. 0.67 m/s          d. 1.0 m/s       e. none of these

19. A physics cart rolls along a low-friction track with considerable momentum. If it rolls at the
same speed but has twice as much mass, its momentum is ____.

a. zero      b. four times as large                  c. twice as large           d. unchanged

20. The firing of a bullet by a rifle causes the rifle to recoil backwards. The speed of the rifle's
recoil is smaller than the bullet's forward speed because the ___.

a. force against the rifle is relatively small       b. speed is mainly concentrated in the bullet
c. rifle has lots of mass                            d. momentum of the rifle is unchanged
e. none of these

21. Two objects, A and B, have the same size and shape. Object A is twice as massive as B. The
objects are simultaneously dropped from a high window on a tall building. (Neglect the effect air
resistance.) The objects will reach the ground at the same time but object A will have a greater
___. Choose all that apply.

a. speed              b. acceleration                              c. momentum
d. none of the above quantities will be greater

22. Cars are equipped with padded dashboards. In collisions, the padded dashboards would be
safer than non-padded ones because they ____. List all that apply.

a. increase the impact time                      b. decrease an occupant's impulse
c. decrease the impact force                     d. none of the above

23. A 4 kg object has a momentum of 12 kg*m/s. The object's speed is ___ m/s.

a. 3         b. 4            c. 12             d. 48            e. none of these.

24. A wad of chewed bubble gum is moving with 1 unit of momentum when it collides with a
heavy box that is initially at rest. The gum sticks to the box and both are set in motion with a
combined momentum that is ___.

a. less than 1 unit           b. 1 unit    c. more than 1 unit            d. not enough information

25. A relatively large force acting for a relatively long amount of time on a relatively small mass
will produce a relatively ______. List all that apply.

a. small velocity change                                        b. large velocity change
c. small momentum change                                        d. small acceleration

26. Consider the recent units on Work and Energy and Momentum and Impulse. Force and time
is related to momentum change in the same manner as force and displacement pertains to
___________.

a. impulse          b. work          c. energy change              d. velocity      e. none of these.

27. A 5-N force is applied to a 3-kg ball to change its velocity from +9 m/s to +3 m/s. This
impulse causes the momentum change of the ball to be ____ kg*m/s.

a. -2.5             b. -10            c. -18           d. -45           e. none of these

28. A 5-N force is applied to a 3-kg ball to change its velocity from +9 m/s to +3 m/s. The
impulse experienced by the ball is ____ N*s.

a. -2.5             b. -10            c. -18           d. -45           e. none of these

29. A 5-N force is applied to a 3-kg ball to change its velocity from +9 m/s to +3 m/s. The
impulse is encountered by the ball for a time of ____ seconds.

a. 1.8        b. 2.5            c. 3.6      d. 10         e. none of these

30. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F.
Assuming the same velocity change of v, the force experienced by a mass of 2M in a time of
(1/2)t is ____.

a. 2F      b. 4F         c. (1/2)*F         d. (1/4)*F           e. none of these

31. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F.
Assuming the same velocity change of v, the force experienced by a mass of 2M in a time of
(1/4)t is ____.

a. 2F      b. 8F         c. (1/2)*F         d. (1/8)*F           e. none of these

32. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F.
Assuming the same velocity change of v, the force experienced by a mass of (1/2)M in a time of
(1/2)t is ____.

a. 2F      b. 4F         c. (1/2)*F         d. (1/4)*F           e. none of these

33. When a mass M experiences a velocity change of v in a time of t, it experiences a force of F.
Assuming the same velocity change of v, the force experienced by a mass of (1/2)M in a time of
4t is ____.

a. 2F      b. 8F         c. (1/2)*F         d. (1/8)*F           e. none of these

34. A 0.50-kg ball moving at 5.0 m/s strikes a wall and rebounds in the opposite direction with a
speed of 2.0 m/s. If the impulse occurs for a time duration of 0.010 s, then the average force
(magnitude only) acting upon the ball is ____ Newtons.

a. 0.14         b. 150           c. 350       d. 500          e. none of these

35. TRUE or FALSE:

If mass and collision time are equal, then impulses are greater on objects which
rebound (or bounce).

a. TRUE                                      b. FALSE

36. Consider the head-on collision between a lady bug and the windshield of a high speed bus.
Which of the following statements are true? List all that apply.

a. The magnitude of the force encountered by the bug is greater than that of the bus.
b. The magnitude of the impulse encountered by the bug is greater than that of the bus.
c. The magnitude of the momentum change encountered by the bug is greater than that
of the bus.
d. The magnitude of the velocity change encountered by the bug is greater than that of
the bus.
e. The magnitude of the acceleration encountered by the bug is greater than that of the
bus.

Answer: D and E

Part C: Diagramming and Analysis

For questions #37-40: Consider the before- and after-collision momentum vectors in the diagram
below. Determine the magnitude and direction of the system momentum before and after the
collision and identify whether or not momentum is conserved. Finally, determine the magnitude
and direction of the net external impulse encountered by the system during the collision.

37.

System Momentum Before Collision:              30 kg•m/s, right
System Momentum After Collision:              30 kg•m/s, right
Is momentum conserved?                     Yes
Net External Impulse During
0N
Collision:
38.

System Momentum Before Collision:     15 kg•m/s, right
System Momentum After Collision:     15 kg•m/s, right
Is momentum conserved?            Yes
Net External Impulse During
0N
Collision:
39.

System Momentum Before Collision:     30 kg•m/s, right
System Momentum After Collision:      10 kg•m/s, left
Is momentum conserved?            No
Net External Impulse During
40 N•s, left
Collision:
40.

System Momentum Before Collision:     50 kg•m/s, right
System Momentum After Collision:     45 kg•m/s, right
Is momentum conserved?            No
Net External Impulse During
5 N•s, left
Collision:
For questions #41-44: Repeat the procedure performed in questions #37-40. Note that these
diagrams give velocity and mass values before and after the collision.

41.

System Momentum Before Collision:             3 kg•m/s, right
System Momentum After Collision:             3 kg•m/s, right
Is momentum conserved?                   Yes
Net External Impulse During
0
Collision:
42.

System Momentum Before Collision:             8 kg•m/s, right
System Momentum After Collision:            10 kg•m/s, right
Is momentum conserved?                   No
Net External Impulse During
2 N•s, right
Collision:
43.

System Momentum Before Collision:            10 kg•m/s, right
System Momentum After Collision:            10 kg•m/s, right
Is momentum conserved?                   Yes
Net External Impulse During
0
Collision:
44.

System Momentum Before Collision:                    0
System Momentum After Collision:                    0
Is momentum conserved?                    Yes
Net External Impulse During
0
Collision:

For questions #45-49, determine the unknown velocity value. Assume that the collisions occur in
an isolated system.

General Note about Q#45-49:

In each of these problems, an expression for the total momentum before the collision is written
by summing the product of mass and velocity for each object. The same is done with the total
momentum after the collision. One must be careful to assign a negative momentum to any object
which is moving leftward. When the velocity of an object is not known, the variable v should be
placed in the expression in place of the actual velocity value. The two expressions for total
momentum before and after the collision are set equal to each other and algebraic operations are
used to determine the unknown velocities.

45.

46.

47.

48.

49.

For questions #50-52, determine the total kinetic energy of the system before and after the
collision and identify the collision as being either perfectly elastic, partially inelastic/elastic or
perfectly inelastic.

50.

Total System Kinetic Energy Before Collision:              50 J
Total System Kinetic Energy After Collision:              50 J
Perfectly Elastic, Partially Inelastic/Elastic or        Perfectly
Perfectly Inelastic?          Elastic
51.

Total System Kinetic Energy Before Collision:               36 J
Total System Kinetic Energy After Collision:               12 J
Perfectly Elastic, Partially Inelastic/Elastic or         Perfectly
Perfectly Inelastic?          Inelastic
52.

Total System Kinetic Energy Before Collision:           183 J
Total System Kinetic Energy After Collision:           153 J
Perfectly Elastic, Partially Inelastic/Elastic or     Partially
Perfectly Inelastic?      Inelastic

Part D: Qualitative Relationships Between Variables

53. An object with a mass M and a velocity v has a momentum of 32 kg•m/s. An object with a
mass of ...

1.   ... 2M and a velocity of 2v would have a momentum of _________.
2.   ... 2M and a velocity of 0.5v would have a momentum of _________.
3.   ... 0.5M and a velocity of 2v would have a momentum of _________.
4.   ... 0.5M and a velocity of 0.5v would have a momentum of _________.
5.   ... 4M and a velocity of v would have a momentum of _________.
6.   ... 4M and a velocity of 0.5v would have a momentum of _________.
7.   ... 0.5M and a velocity of 4v would have a momentum of _________.
8.   ... 3M and a velocity of 2v would have a momentum of _________.

1.   ... 2M and a velocity of 2v would have a momentum of 64 kg•m/s.
2.   ... 2M and a velocity of 0.5v would have a momentum of 32 kg•m/s.
3.   ... 0.5M and a velocity of 2v would have a momentum of 32 kg•m/s.
4.   ... 0.5M and a velocity of 0.5v would have a momentum of 8 kg•m/s.
5.   ... 4M and a velocity of v would have a momentum of 128 kg•m/s.
6.   ... 4M and a velocity of 0.5v would have a momentum of 64 kg•m/s.
7.   ... 0.5M and a velocity of 4v would have a momentum of 64 kg•m/s.
8.   ... 3M and a velocity of 2v would have a momentum of 192 kg•m/s.

54. Two carts are placed next to each other on a low-friction track. The carts are equipped with a
spring-loaded mechanism which allows them to impart an impulse to each other. Cart A has a
mass of M and Cart B has a mass of M. The spring-loaded mechanism is engaged and then
released. The impulse causes Cart A to be propelled forward with a velocity of 40 cm/s.

a. Cart B will be propelled backward with a velocity of ____________.
b. ... If Cart B had a mass of 2M then it would be propelled backwards with a velocity
of ____________..
c. ... If Cart B had a mass of 0.5M then it would be propelled backwards with a velocity
of ____________.
d. ... If Cart B has a mass of 2M then it would be propelled backwards with a
momentum which is ____________ original momentum.
e. ... If Cart B has a mass of 2M then it would encounter an impulse which is
____________ original impulse.

a. Cart B will be propelled backward with a velocity of 40 cm/s.
b. ... If Cart B had a mass of 2M then it would be propelled backwards with a velocity of 20
cm/s.
c. ... If Cart B had a mass of 0.5M then it would be propelled backwards with a velocity of
80 cm/s.
d. ... If Cart B has a mass of 2M then it would be propelled backwards with a momentum
which is the same as the original momentum.
e. ... If Cart B has a mass of 2M then it would encounter an impulse which is the same as
the original impulse.

Part E: Problem-Solving

55. A 0.530-kg basketball hits a wall head-on with a forward speed of 18.0 m/s. It rebounds with
a speed of 13.5 m/s. The contact time is 0.100 seconds. (a) determine the impulse with the wall,
(b) determine the force of the wall on the ball.

Answer: (a) -16.7 N s; (b) -167 N

56. A 4.0-kg object has a forward momentum of 20.0 kg•m/s. A 60.0 N•s impulse acts upon it in
the direction of motion for 5.0 seconds. A resistive force of 6.0 N then impedes its motion for 8.0
seconds. Determine the final velocity of the object.

Answer: vf = 8.0 m/s

57. A 3.0-kg object is moving forward with a speed of 6.0 m/s. The object then encounters a
force of 2.5 N for 8.0 seconds in the direction of its motion. The object then collides head-on
with a wall and heads in the opposite direction with a speed of 5.0 m/s. Determine the impulse
delivered by the wall to the object.

58. A 46-gram tennis ball is launched from a 1.35-kg homemade cannon. If the cannon recoils
with a speed of 2.1 m/s, determine the muzzle speed of the tennis ball.

59. A 2.0-kg box is attached by a string to a 5.0-kg box. A compressed spring is placed between
them. The two boxes are initially at rest on a friction-free track. The string is cut and the spring
applies an impulse to both boxes, setting them in motion. The 2.0-kg box is propelled backwards
and moves 1.2 meters to the end of the track in 0.50 seconds. Determine the time it takes the 5.0-
kg box to move 0.90 meters to the opposite end of the track.

60. A 2.80-kg physics cart is moving forward with a speed of 45.0 cm/s. A 1.90-kg brick is
dropped from rest and lands on the cart. The cart and brick move together across the horizontal
surface. Assume an isolated system.

a. Determine the post-collision speed of the cart and the brick.

b. Determine the momentum change of the cart.

c. Determine the momentum change of the brick.

d. Determine the net impulse upon the cart.

e. Determine the net impulse upon the system of cart and brick.

Answers: (a) v = 26.8 cm/s

(b)  pcart = -50.9 kg • cm/s

(c) pbrick = +50.9 kg • cm/s

(d) Impulse on cart = -50.9 kg • cm/s

(e) Impulse on brick = +50.9 kg • cm/s

61. Two ice skaters collide on the ice. A 39.6-kg skater moving South at 6.21 m/s collides with a
52.1-kg skater moving East at 4.33 m/s. The two skaters entangle and move together across the
ice. Determine the magnitude and direction of their post-collision velocity.

Answer: 3.64 m/s at 42.5 degrees east of south (312.5 degrees)
ENERGY REVIEW
Part A: Forced Choice Questions

1. Which of the following statements are true about work? Include all that apply.

a.   Work is a form of energy.
b.   A Watt is the standard metric unit of work.
c.   Units of work would be equivalent to a Newton times a meter.
d.   A kg•m2/s2 would be a unit of work.
e.   Work is a time-based quantity; it is dependent upon how fast a force displaces an object.
f.   Superman applies a force on a truck to prevent it from moving down a hill. This is an
example of work being done.
g.   An upward force is applied to a bucket as it is carried 20 m across the yard. This is an
example of work being done.
h.   A force is applied by a chain to a roller coaster car to carry it up the hill of the first drop
of the Shockwave ride. This is an example of work being done.
i.   The force of friction acts upon a softball player as she makes a headfirst dive into third
base. This is an example of work being done.
j.   An eraser is tied to a string; a person holds the string and applies a tension force as the
eraser is moved in a circle at constant speed. This is an example of work being done.
k.   A force acts upon an object to push the object along a surface at constant speed. By itself,
this force must NOT be doing any work upon the object.
l.   A force acts upon an object at a 90-degree angle to the direction that it is moving. This
force is doing negative work upon the object.
m.   An individual force does NOT do positive work upon an object if the object is moving at
constant speed.
n.   An object is moving to the right. A force acts leftward upon it. This force is doing
negative work.
o.   A non-conservative force is doing work on an object; it is the only force doing work.
Therefore, the object will either gain or lose mechanical energy.

2. Which of the following statements are true about power? Include all that apply.

a. Power is a time-based quantity.
b. Power refers to how fast work is done upon an object.
c. Powerful people or powerful machines are simply people or machines which always do a
lot of work.
d. A force is exerted on an object to move it at a constant speed. The power delivered by
this force is the magnitude of the force multiplied by the speed of the object.
e. The standard metric unit of power is the Watt.
f. If person A and person B do the same job but person B does it faster, then person A does
more work but person B has more power.
g. The Newton•meter is a unit of power.
h. A 60-kg boy runs up a 2.0 meter staircase in 1.5 seconds. His power is approximately 80
Watt.
i. A 300-Newton force is applied to a skier to drag her up a ski hill at a constant speed of
1.5 m/s. The power delivered by the toe rope is 450 Watts.

3. Consider the following physical situations. For each case, determine the angle between the
indicated force (in boldface type) and the displacement ("theta" in the work equation).

A. 0          B. 180           C. 90          D. 30           E. 60
DEGREES        DEGREES         DEGREES        DEGREES         DEGREES

a. A rightward applied force is used to displace a television set to the right.
b. The force of friction acts upon a rightward-moving car to bring it to a stop. A waiter uses
an applied force to balance the weight of a tray of plates as he carries the tray across the
room.
c. The force of air resistance acts upon a vertically-falling skydiver. The force of friction
acts upon a baseball player as he slides into third base. An applied force is used by a
freshman to lift a World Civilization book to the top shelf of his locker.
d. A bucket of water is tied to a string and tension supplies the centripetal force to keep it
moving in a circle at constant speed. An applied force acting at 30-degrees to the
horizontal is used to displace an object to the right.
e. A group of football players use an applied force to push a sled across the grass.
f. The tension in the elevator cable causes the elevator to rise at a constant speed.
g. In a physics lab, an applied force is exerted parallel to a plane inclined at 30-degrees in
order to displace a cart up the incline. An applied force is exerted upwards and
rightwards at an angle of 30-degrees to the vertical in order to displace an object to the
right.
h. A child rests on the seat of a swing which is supported by the tension in its cables; he
swings from the highest position to its lowest position.

a.   A - 0 degrees
b.   B - 180 degrees
c.   C - 90 degrees
d.   B - 180 degrees
e.   B - 180 degrees
f.   A - 0 degrees
g.   C - 90 degrees
h.   D - 30 degrees
i.   A - 0 degrees
j.   A - 0 degrees
k.   A - 0 degrees
l.   E - 60 degrees
m.   C - 90 degrees

4. Consider the following physical situations. Identify whether the indicated force (in boldface
type) does positive work, negative work or no work.

A. Positive Work        B. Negative Work         C. No Work

Description of Physical Situation                                             +, -, or no Work
a. A cable is attached to a bucket and the force of tension is used to pull
the bucket out of a well.
b. Rusty Nales uses a hammer to exert an applied force upon a stubborn
nail to drive it into the wall.
c. Near the end of the Shockwave ride, a braking system exerts an
applied force upon the coaster car to bring it to a stop.
d. The force of friction acts upon a baseball player as he slides into
third base.
e. A busy spider hangs motionless from a silk thread, supported by the
tension in the thread.
f. In baseball, the catcher exerts an abrupt applied force upon the ball to
stop it in the catcher's mitt.
g. In a physics lab, an applied force is exerted parallel to a plane
inclined at 30-degrees in order to displace a cart up the incline.
h. A pendulum bob swings from its highest position to its lowest
position under the influence of the force of gravity.

Answer: See table below; explanations provided below.

Description of Physical Situation                                             +, -, or no Work
a. A cable is attached to a bucket and the force of tension is used to pull
A. Positive Work
the bucket out of a well.
b. Rusty Nales uses a hammer to exert an applied force upon a stubborn
A. Positive Work
nail to drive it into the wall.
c. Near the end of the Shockwave ride, a braking system exerts an
B. Negative Work
applied force upon the coaster car to bring it to a stop.
d. The force of friction acts upon a baseball player as he slides into
B. Negative Work
third base.
e. A busy spider hangs motionless from a silk thread, supported by the
C. No Work
tension in the thread.
f. In baseball, the catcher exerts an abrupt applied force upon the ball to
B. Negative Work
stop it in the catcher's mitt.
g. In a physics lab, an applied force is exerted parallel to a plane
A. Positive Work
inclined at 30-degrees in order to displace a cart up the incline.
h. A pendulum bob swings from its highest position to its lowest
A. Positive Work
position under the influence of the force of gravity.

5. Which of the following statements are true about conservative(internal) and non-conservative
(external) forces? Include all that apply.

a. A force is regarded as a conservative force if it does work but does not remove
mechanical energy from a system of objects.
b. A force is regarded as a non-conservative force if it does not add mechanical energy to a
system of objects.
c. The force of gravity and elastic (spring) force are both examples of a conservative forces.
d. Applied forces, air resistance, friction forces, and tension are common examples of non-
conservative forces.
e. Physicists envy biologists' ability to instill order on the world of animal species through
their taxonomic system. So physicists have made a habit of identifying forces as
conservative and non-conservative forces in order to instill order on the world of forces.
f. If a non-conservative force acts upon an object, then the object will either gain or lose
mechanical energy.
g. If the only forces which do work upon an object are conservative forces, then the object
will conserve its mechanical energy.
h. If the sum of an object's KE and PE is remaining constant, then non-conservative forces
are NOT doing work.
i. If work is NOT done on an object by a non-conservative force, then the object will
experience a transformation of energy from kinetic to potential energy (or vice versa).
j. An object starts from an elevated position with 50 J of potential energy and begins its fall
towards the ground. If non-conservative forces can be assumed to NOT do work, then at
some point during the fall the object will have 20 J of potential energy and 30 J of kinetic
energy.

Answer: A(sort of) CDGH I(sort of) J

6. Which of the following statements are true about kinetic energy? Include all that apply.

a. Kinetic energy is the form of mechanical energy which depends upon the position of an
object.
b. If an object is at rest, then it does not have any kinetic energy.
c.   If an object is on the ground, then it does not have any kinetic energy.
d.   The kinetic energy of an object is dependent upon the weight and the speed of an object.
e.   Faster moving objects always have a greater kinetic energy.
f.   More massive objects always have a greater kinetic energy.
g.   Kinetic energy is a scalar quantity.
h.   An object has a kinetic energy of 40 J. If its mass were twice as much, then its kinetic
energy would be 80 J.
i.   An object has a kinetic energy of 40 J. If its speed were twice as much, then its kinetic
energy would be 80 J.
j.   Object A has a mass of 1 kg and a speed of 2 m/s. Object B has a mass of 2 kg and a
speed of 1 m/s. Objects A and B have the same kinetic energy.
k.   An object can never have a negative kinetic energy.
l.   A falling object always gains kinetic energy as it falls.
m.   A 1-kg object is accelerated from rest to a speed of 2.0 m/s. This object gains 4.0 Joules
of kinetic energy.
n.   If work is done on an object by a non-conservative force, then the object will either gain
or lose kinetic energy.

7. Which of the following statements are true about mechanical energy? Include all that apply.

a. The total amount of mechanical energy of an object is the sum of its potential energy and
the kinetic energy.
b. Heat is a form of mechanical energy.
c. The mechanical energy of an object is always conserved.
d. When non-conservative forces do work, energy is transformed from kinetic to potential
(or vice versa), but the total mechanical energy is conserved.
e. A bowling ball is mounted from a ceiling by way of a strong cable. It is drawn back and
released, allowed to swing as a pendulum. As it swings from its highest position to its
lowest position, the total mechanical energy is mostly conserved.
f. When a friction force does work on an object , the total mechanical energy of that object
is changed.
g. The total mechanical energy of an object remains constant if the only forces doing work
on the object are conservative forces.
h. If an object gains mechanical energy, then one can be certain that a non-conservative
force is doing work.

8. Rank these four objects in increasing order of potential energy, beginning with the smallest.

Object A                  Object B                 Object C                  Object D

m = 5.0 kg               m = 10.0 kg               m = 1.0 kg               m = 5.0 kg

v = 4.0 m//s             v = 2.0 m//s              v = 5.0 m/s              v = 2.0 m//s

h = 2.0 m               h = 3.00 m                h = 5.0 m                 h = 4.0 m

Answer: C < A < D < B

NOTE: The next 15 questions presume that the value of g is 10 m/s/s.

9. A 1200 kg car and a 2400 kg car are lifted to the same height at a constant speed in a auto
service station. Lifting the more massive car requires ____ work.

a. less work                    b. the same                                c. twice as much
d. four times as much           e. more than 4 times as much

10. An arrow is drawn back so that 50 Joules of potential energy is stored in the stretched bow
and string. When released, the arrow will have a kinetic energy of ____ Joules.

a. 50     b. more than 50          c. less than 50

11. A child lifts a box up from the floor. The child then carries the box with constant speed to the
other side of the room and puts the box down. How much work does he do on the box while
walking across the floor at constant speed?

a. zero J                    b. more than zero J

12. A 1000-kg car is moving at 40 km/hr when the driver slams on the brakes and skids to a stop
(with locked brakes) over a distance of 20 meters. How far will the car skid with locked brakes if
it is traveling at 120 km/hr?

a. 20 m             b. 60 m.            c. 90 m           d. 120 m               e. 180 m

13. A platform diver weighs 500 N. She steps off a diving board that is elevated to a height of 10
meters above the water. The diver will possess ___ Joules of kinetic energy when she hits the
water.

a. 10      b. 500          c. 510        d. 5000           e. more than 5000 .

14. A ball is projected into the air with 100 J of kinetic energy. The kinetic energy is transformed
into gravitational potential energy on the path towards the peak of its trajectory. When the ball
returns to its original height, its kinetic energy is ____ Joules. Do consider the effects of air
resistance

a. less than 100                         b. 100           c. more than 100
d. not enough information given

15. During a construction project, a 2500 N object is lifted high above the ground. It is released
and falls 10.0 meters and drives a post 0.100 m into the ground. The average impact force on the
object is ____ Newtons.

a. 2500             b. 25000              c. 250,000                 d. 2,500,000

16. A 10-Newton object moves to the left at 1 m/s. Its kinetic energy is approximately ____
Joules.

a. 0.5              b. 1        c. 10              d. more than 10

17. Luke Autbeloe stands on the edge of a roof throws a ball downward. It strikes the ground
with 100 J of kinetic energy. Luke now throws another identical ball upward with the same
initial speed, and this too falls to the ground. Neglecting air resistance, the second ball hits the
ground with a kinetic energy of ____ J.

a. less than 100           b. 100     c. 200     d. more than 200             e. none of these

18. An object at rest may have __________.

a. speed       b. velocity          c. acceleration          d. energy           e. all of these

19. A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height
(relative to the water) from which the diver dove was approximately ____ meters.

a. 5               b. 10                       c. 50                    d. 100

20. A job is done slowly, and an identical job is done quickly. Both jobs require the same amount
of ____, but different amounts of ____. Pick the two words which fill in the blanks in their
respective order.

a. energy, work        b. power, work           c. work, energy         d. work, power

e. power, energy       f. force, work           g. power, force         h. none of these

21. Which requires more work: lifting a 50 kg crate a vertical distance of 2 meters or lifting a 25
kg crate a vertical distance of 4 meters?

a. lifting the 50 kg crate                                        b. lifting the 25 kg crate
c. both require the same amount of work

Part B: Straightforward Computational Problems

22. Approximate the work required lift a 2.5-kg object to a height of 6.0 meters. PSYW

23. In the It's All Uphill Lab, a force of 20.8 N is applied parallel to the incline to lift a 3.00-kg
loaded cart to a height of 0.450 m along an incline which is 0.636-m long. Determine the work
done upon the cart and the subsequent potential energy change of the cart. PSYW

24. A 50-kg hiker ascends a 40-meter high hill at a constant speed of 1.2 m/s. If it takes 400 s to
climb the hill, then determine ... . PSYW

a.   kinetic energy change of the hiker.
b.   the potential energy change of the hiker.
c.   the work done upon the hiker.
d.   the power delivered by the hiker.

a.   (Delta = ) KE = 0 J
b.   (Delta) PE = +20000 J
c.   W = +20000 J
d.   P = 50 Watts

25. An 800.0-kg car skids to a stop across a horizontal surface over a distance of 45.0 m. The
average force acting upon the car is 7000.0 N. Determine ... . PSYW

a.   the work done upon the car.
b.   the initial kinetic energy of the car.
c.   the acceleration of the car.
d.   the initial velocity of the car.

a.   W = -315000 J
b.   KEi = +315000 J
c.   a = -8.75 m/s/s
d.   vi = 28.1 m/s

26. A 60.0-kg skier accelerates down an icy hill from an original height of 500.0 meters. Use the
work-energy theorem to determine the speed at the bottom of the hill if...

a. ... no energy is lost or gained due to friction, air resistance and other non-conservative forces.
PSYW

b. ... 1.40 x 105 J of energy are lost due to external forces. PSYW

Answers: (a) v = 1.00 x 102 m/s; (b) v = 73.0 m/s
27. Use the work-energy theorem to determine the force required to stop a 1000.0-kg car moving
at a speed of 20.0 m/s if there is a distance of 45.0 m in which to stop it. PSYW

Answer: F = 4.44*103 N

Part C: Work-Energy Bar Charts, Analysis, and Conceptual Reasoning

28. Consider the following physical situations. Identify the forces which do work upon the
indicated object (in boldface type) and categorize them as conservative or non-conservative
forces. Then indicate whether the total mechanical energy of the object changes; it it changes,
then indicate whether the change is a positive or negative change. Finally, indicate whether the
potential energy and the kinetic energy changes; if PE or KE changes, then indicate whether the
change is a positive or negative change.

Identity of Forces
Change in    Change in     Change in
TME           PE           KE
Which Do Work:
Description of Physical                     External
Yes           Yes            Yes
Situation                            Internal
(Non-             No             No             No
(Conserv)              (+ or        (+ or -        (+ or -
-)            )              )
Conserv)
a. A force is applied to move a
physics cart from the floor to
the top of an inclined plane at a
constant speed.
b. A physics student scurries
up a flight of stairs at constant
speed.
c. In a moment of unsupervised
phun, a physics student hoists
herself onto a staircase banister
and accelerates down the
banister. Ignore all friction
forces.
d. A ball is dropped from rest
from on the top of a hill and
falls to the ground below.
Ignore air resistance.
e. A ball leaves top of a hill
with a large horizontal velocity.
It falls to the ground below.
Ignore air resistance.
f. A Hot Wheels car is at rest at
an elevated position along an
inclined plane; it is released and
rolls to a position along the
ground. Ignore air resistance.
g. A Hot Wheels car is in
motion at the bottom of a hill
when it hits a computer diskette
box and skids to a stop.
h. A pendulum bob swings
from its highest position to its
lowest position.
i. A physics cart is released
from rest at an elevated position
along an inclined plane; it is
released and rolls to a position
along the incline approximately
5 cm from the bottom.

Identity of Forces
Change in    Change in     Change in
TME           PE           KE
Which Do Work:
Description of Physical                      External
Yes           Yes            Yes
Situation                             Internal
(Non-             No             No             No
(Conserv)              (+ or        (+ or -        (+ or -
-)            )              )
Conserv)
a. A force is applied to move a
physics cart from the floor to
Gravity    Applied     +             +                      No
the top of an inclined plane at a
constant speed.
b. A physics student scurries                    Applied
up a flight of stairs at constant     Gravity      or        +             +                      No
speed.                                           Normal
c. In a moment of unsupervised        Gravity     None              No      -             +
phun, a physics student hoists
herself onto a staircase banister
and accelerates down the
banister. Ignore all friction
forces.
d. A ball is dropped from rest
from on the top of a hill and
Gravity    None              No        -              +
falls to the ground below.
Ignore air resistance.
e. A ball leaves top of a hill
with a large horizontal velocity.
Gravity    None              No        -              +
It falls to the ground below.
Ignore air resistance.
f. A Hot Wheels car is at rest at
an elevated position along an
inclined plane; it is released and Gravity     None              No        -              +
rolls to a position along the
ground. Ignore air resistance.
g. A Hot Wheels car is in
motion at the bottom of a hill
None     Applied     -                      No       -
when it hits a computer diskette
box and skids to a stop.
h. A pendulum bob swings
from its highest position to its    Gravity    None              No        -              +
lowest position.
i. A physics cart is released
from rest at an elevated position
along an inclined plane; it is
Gravity      None              No        -              +
released and rolls to a position
along the incline approximately
5 cm from the bottom.
Answer: See table above

Note that whenever a non-conservative force is doing work (as in a, b, g), the total mechanical
energy is changing.

However, when the only forces which do work are conservative forces, the total mechanical
energy will remain constant (i.e., be conserved). In such cases, energy will change form from
potential to kinetic energy (or vice versa); yet the total amount of the two forms would not be
changed.

Potential energy can be considered to change if an object changes its height; if the height
decreases, then the potential energy decreases; it the height increases, then the potential energy
increases.

Kinetic energy can be considered to change if an object changes its speed; if the speed decreases,
then the kinetic energy decreases; it the speed increases, then the kinetic energy increases.

29. Several physical situations are described below. For each situation, simplify the work-energy
equation by canceling any zero terms and any energy terms (whether KE or PE) which are
unchanging. Explain each term which gets canceled. The first problem is done as an example.

Simplification of Work-Energy
Description of Physical Situation
Equation
a. A ball starts from rest on top of
a tall pillar and falls to the ground
below. Assume the effect of air
resistance is negligible.

b. A ball starts from
rest at an elevated
position along an
inclined plane and
rolls to the ground
below. Assume that
the effect of friction
and air resistance is
negligible.
c. A ball starts from
rest at an elevated
position along an
inclined plane and
rolls to the ground
below. Consider the
effect of friction and
air resistance.

d. A track is constructed by
stretching a grooved and
pliable material between two
lab poles. A metal ball starts
from rest at point A and rolls
to point B. Friction and air
resistance have an effect upon the ball's motion.

e. A pendulum bob is
mounted on top of a
lab pole and drawn
back to a string which
it tied between two
other poles. The pendulum bob is released from rest.
Upon reaching the vertical, the string hits a barrier and
a new pivot point is established as the bob continues in
motion along an upward trajectory. Ignore the effect of
air resistance.
f. A Hot Wheels car starts
from rest on top of an
inclined plane and rolls down
the incline through a loop
and along a horizontal
surface. Friction and air
resistance have a significant effect on the car.
g. An unattended hot dog wagon
starts from rest and rolls down a
hill and up a second hill. Ignore
the effect of friction and air
resistance.

h. A roller coaster car is
already in motion on the
top of the first drop and
rolls along the track over
a couple of hills. Ignore
the effect of friction and air resistance.

i. A cross-
country skier
is in motion
on top of a
small hill. He
skis down the
hill into the valley and up a second smaller hill. He
uses his poles to propel himself during the entire
motion. Ignore the effect of friction and air resistance.

Simplification of Work-Energy
Description of Physical Situation
Equation
a. A ball starts from rest on top of               PEi = KEf
a tall pillar and falls to the ground
below. Assume the effect of air              (Since initially at rest,
resistance is negligible.                     KEi = 0 and cancels.
Since the final height is
0, PEf = 0 and cancels.
Since non-conservative
forces are not doing
work, Wnc = 0)
b. A ball starts from                    PEi = KEf
rest at an elevated
position along an        (Initially the ball is at rest, KEi = 0 and
inclined plane and       cancels. Since the final height is 0, PEf
rolls to the ground      = 0 and cancels. Since non-conservative
below. Assume that         forces are not doing work, Wnc = 0)
the effect of friction
and air resistance is
negligible.

c. A ball starts from                PEi + Wnc = KEf
rest at an elevated
position along an        (Initially the ball is at rest, KEi = 0 and
inclined plane and       cancels. Since the final height is 0, PEf
rolls to the ground         = 0 and cancels. Friction is a non-
below. Consider the      conservative forces and it does work so
effect of friction and             Wnc does not cancel.)
air resistance.

d. A track is constructed by                  PEi + Wnc = PEf
stretching a grooved and
pliable material between two      (The ball is at rest in both its initial and
lab poles. A metal ball starts   final states, so both KEi = 0 and KEf = 0
from rest at point A and rolls    and cancels. Friction and air resistance
to point B. Friction and air     are non-conservative forces and do work
resistance have an effect upon the ball's motion.                   so Wnc does not cancel.)
e. A pendulum bob is                  PEi = KEf + PEf
mounted on top of a
lab pole and drawn       (Initially the ball is at rest, KEi = 0 and
back to a string which      cancels. At the point shown in the
it tied between two       diagram as the final state, the ball has
other poles. The pendulum bob is released from rest.      both height and would still be moving;
Upon reaching the vertical, the string hits a barrier and    so neither final energy term would
a new pivot point is established as the bob continues in cancel. The only non-conservative force
motion along an upward trajectory. Ignore the effect of     present - tension - does not do work
air resistance.                                           since it is directed at 90-degrees to the
direction of motion; so Wnc cancels.)
f. A Hot Wheels car starts                PEi + Wnc = KEf
from rest on top of an
inclined plane and rolls down (Initially the car is at rest, KEi = 0 and
the incline through a loop    cancels. Since the final height is 0, PEf
and along a horizontal          = 0 and cancels. Friction is a non-
surface. Friction and air     conservative forces and it does work so
resistance have a significant effect on the car.                   Wnc does not cancel.)

g. An unattended hot dog wagon                    PEi = KEf + PEf
starts from rest and rolls down a
hill and up a second hill. Ignore       (Initially the wagon is at rest, KEi = 0
the effect of friction and air        and cancels. The wagon has height in its
resistance.                              initial state, so the PEi term does not
cancel. At the point shown in the
diagram as the final state, the wagon has
both height and would still be moving;
so neither final energy term would
cancel. There is no friction nor air
resistance and the normal force does not
do work since it is perpendicular to the
displacement; so Wnc cancels.)
h. A roller coaster car is                PEi = KEf + PEf
already in motion on the
top of the first drop and   (Initially the car is at rest, KEi = 0 and
rolls along the track over cancels. The car has height in its initial
a couple of hills. Ignore state, so the PEi term does not cancel. At
the effect of friction and air resistance.                   the point shown in the diagram as the
final state, the car has both height and
would still be moving; so neither final
energy term would cancel. There is no
friction nor air resistance and the normal
force does not do work since it is
perpendicular to the displacement; so
Wnc cancels.)
i. A cross-         KEi + PEi + Wnc = KEf + PEf
country skier
is in motion      (Initially, the skier has height and
on top of a      motion and so neither initial energy
small hill. He term will cancel. The skier is using his
skis down the poles to propel himself so there is a non-
hill into the valley and up a second smaller hill. He     conservative force doing work; the Wnc
uses his poles to propel himself during the entire        term does not cancel. In the final state,
motion. Ignore the effect of friction and air resistance.    the skier is moving and has height
(presuming that the zero level is the
valley) and so neither final energy term
will cancel.

Part D: Complex Analysis and Problem-Solving

30. A 21.3-kg child positions himself on an inner-tube which is suspended by a 7.28-m long rope
attached to a strong tree limb. The child and tube are drawn back until it makes a 17.4-degree
angle with the vertical. The child is released and allowed to swing to and from. Assuming
negligible friction, determine the child's speed at his lowest point in the trajectory. PSYW

31. A baseball player catches a 163-gram baseball which is moving horizontally at a speed of
39.8 m/s. Determine the force which she must apply to the baseball if her mitt recoils a
horizontal distance of 25.1 cm. PSYW

32. A 62.9-kg downhill skier is moving with a speed of 12.9 m/s as he starts his descent from a
level plateau at 123-m height to the ground below. The slope has an angle of 14.1 degrees and a
coefficient of friction of 0.121. The skier coasts the entire descent without using his poles; upon
reaching the bottom he continues to coast to a stop; the coefficient of friction along the level
surface is 0.623. How far will he coast along the level area at the bottom of the slope? PSYW

33. A 221-gram ball is thrown at an angle of 17.9
degrees and a speed of 36.7 m/s from the top of a 39.8-m
high cliff. Determine the impact speed of the ball when it
strikes the ground. Assume negligible air resistance.
PSYW

34. Claire deAisles has just completed her shopping at the Jewel
food store. She accidentally bumps her 42.5-kg cart, setting it in
motion from rest down a hill inclined at 14.9 degrees. Upon
descending a distance of 9.27 meters along the inclined plane, the
cart hits a tree stump (which was placed in the parking lot for the
sole purpose of this problem). A 0.295-kg can of tomato soup is immediately hurled from the
moving cart and heads towards Will N. Tasue's brand new Lexus. Upon striking the Lexus, the
tomato soup can creates a dent with a depth of 3.16 cm. Noah Formula, who is watching the
entire incident and fixing to do some physics, attempts to calculate the average force which the
Lexus applies to the soup can. Assume negligible air resistance and friction forces and help Noah
out. PSYW

35. Mia Kneezhirt jumps from a second story dorm room (h = 7.91 m) to the ground below.
Upon contact with the ground, she allows her 62.4-kg body to come to an abrupt stop as her
center of gravity is displaced downwards a distance of 89.2 cm. Calculate the average upward
force exerted by the ground upon Mia's fragile body. PSYW (Use Conservation of Energy
Equation)

Answer: 5.42 X 103 N

36. In a physics lab, a 0.500-kg cart moving at 36.4 cm/s collides inelastically with a second cart
which is initially at rest. The two carts move together with a speed of 21.8 cm/s after the
collision. Determine the mass of the second cart.

Answer: ~ 0.335 kg

37. A classic physics demonstration involves firing a bullet into a block of wood suspended by
strings from the ceiling. The height to which the wood rises below its lowest position is
mathematically related to the pre-collision speed of the bullet. If a 10.3-gram bullet is fired into
the center of a 1.5-kg block of wood and it rises upward a distance of 25 cm, then what was the
pre-collision speed of the bullet?

Answer: 3.2 x 102 m/s
38. On Okanagan Lake, Mr. MacKenzie (108 kg) and his brother (92 kg) are cruising around in
Ogopogo shaped boats (160 kg). Mr. MacKenzie’s gets pedaling at 1.0 m/s and collides head on
with his brother who is moving in the opposite direction at 1.2 m/s. After the collision, his
brother bounces backwards at 0.30 m/s. Assuming an isolated system, determine

a. ... Mr. MacKenzie's post-collision speed.

b. ... the percentage of original kinetic energy which is lost as the result of the collision

Answer: (a) v = ~0.41 m/s

(b) % KE Loss = ~89%

39. Two carbon (ultra hard and virtually undeformable) cars collide elastically on a level, low-
friction track. Car A has a mass of 1675 kg and is moving east at 15.0 m/s. Car B has a mass of
1350 kg and is moving West at 12.0 m/s. Determine the post-collision velocities of the two cars.

Answer: vA-after = -9.10 cm/s; vB-after = 17.9 cm/s

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