Download AP QUIZ #13 INTRO TO ENERGY

AP QUIZ #13 INTRO TO ENERGY
1) CHANGE IN KINETIC ENERGY
A cart with a spring plunger runs into a fixed barrier. The mass of the cart, its velocity just before impact with the
barrier, and its velocity right after collision are given in each figure. (All velocities before the collision are given as
positive since the cart is moving to the right. After the collision the cart is either moving to the left, indicated with a
negative velocity, or is at rest.)
A
C
Before
After
10 kg
vo = 4 m/s
Before
After
10 kg
10 kg
10 kg
vf = 0
vo = 3 m/s
Before
After
20 kg
20 kg
vo = 1 m/s
B
D
vf = –1 m/s
vf = –1 m/s
Before
After
5 kg
5 kg
vo = 5 m/s
vf = –3 m/s
Rank the change in kinetic energy for each cart in these situations.
OR
1
Greatest
2
3
4
Least
All
the same
All
zero
Cannot
determine
Explain your reasoning.
Answer: A > B = D > C. The change in kinetic energy is the final minus the initial which gives the sequence here
since kinetic energy is a scalar and we are only interested in the change.
2) EQUAL FORCES ON BOXES—WORK DONE ON BOX
In the figures below, identical boxes of mass 10 kg are moving at the same initial velocity to the right on a flat
surface. The same magnitude force, F, is applied to each box for the distance, d, indicated in the figures.
A
B
C
F
d=5m
D
F
F
d=5m
d = 10 m
E
F
F
F
F
d = 10 m
d=5m
d=5m
Rank these situations in order of the work done on the box by F while the box moves the indicated distance to
the right.
OR
1
Greatest
2
3
4
5
6
Least
All
the same
All
zero
Cannot
determine
Explain your reasoning.
Answer: B > A > C > D = F > E. The work done on the box is given by the product of the component of the force
in the direction of motion times the distance moved. Positive work is done in A, B, C, zero work in D and F since
the force and displacement are perpendicular to each other, and negative work in E, i.e., the box does work on
the agent exerting F rather than work being done on the box.
3) VELOCITY TIME GRAPH I—WORK DONE ON BOX
Shown below is a graph of velocity versus time for an object that moves along a straight, horizontal line under the,
perhaps intermittent, action of a single force exerted by an external agent.
Velocity (m/s)
8
6
4
2
B
C
A
D
0
5
10
15
20 Time (s)
–2
–4
–6
–8
Rank the 5-second intervals shown on the graph on the work done on the box by the external agent.
OR
1
Greatest
2
3
4
Least
All
the same
All
zero
Cannot
determine
Explain your reasoning.
Answer: B > A = D > C. In this situation the work done on the box will change its kinetic energy, so to
find how much work was done we need to subtract the initial kinetic energy from the final value for
each 5 second interval. The external agent will do positive work in interval B, no work in intervals A
and D since the kinetic energy doesn’t change in those intervals, and negative work in interval C, i.e.,
the box is doing work on the external agent in C rather than the agent doing work on the box.
4) VELOCITY TIME GRAPH II—WORK DONE ON BOX
Shown below is a graph of velocity versus time for an object that moves along a straight, horizontal line under the,
perhaps intermittent, action of a single force exerted by an external agent.
Velocity (m/s)
8
6
4
2
0
A
5
10
15
–2
D
20 Time (s)
B
–4
–6
C
–8
Rank the 5-second intervals shown on the graph on the work done on the box by the external agent.
OR
1
Greatest
2
3
4
Least
All
the same
All
zero
Cannot
determine
Explain your reasoning.
Answer: D > A = C > B. In this situation the work done on the box will change its kinetic energy, so to find how
much work was done we need to subtract the initial kinetic energy from the final value for each 5 second interval.
The external agent will do positive work in interval D, no work in intervals A, and C since the kinetic energy
doesn’t change in those intervals, and negative work in interval B, i.e., the box does work on the external agent in
B rather than the agent doing work on the box.
5) FORCE PUSHING BOX—CHANGE IN KINETIC ENERGY
Various similar boxes are being pushed for 10 m across a floor by a net horizontal force as shown below. The mass
of the boxes and the net horizontal force for each case are given in the indicated figures. All boxes have an initial
velocity of 10 m/s to the right.
A
B
F = 100 N
C
F = 50 N
10 kg
D
F = 75 N
15 kg
F = 100 N
20 kg
15 kg
Rank the change in kinetic energy for each box from the greatest change in kinetic energy to the least change
in kinetic energy.
OR
1
Greatest
2
3
4
Least
All
the same
All
zero
Cannot
determine
Explain your reasoning.
Answer: A = D > C > B. The change in kinetic energy will occur because of the energy transfer produced by the
external agent working on the box, so, since all of the boxes move the same distance under the action of the
external force, the magnitudes of the external forces determine the ranking.
6) BOAT POSITION VS. TIME GRAPHS—WORK
Shown are graphs of the position versus time for two boats traveling along a narrow channel. The scales on both
axes are the same for the graphs. In each graph, two points are marked with dots.
A Position
B Position
Time
Time
A student who is using these graphs to compare the net work done on the two boats between the two points says:
“I think that more net work was done on the boat in graph B because it moved farther during the
interval between the points.”
What, if anything, is wrong with this statement? If something is wrong, identify it, and explain how to correct
it. If this statement is correct, explain why.
Answer: The student’s contention is wrong. Since both graphs have a straight line for the motion of the sailboats,
they moved at constant speeds, so there was no change in their kinetic energy and, consequently, no work was
done on them.
7) TUGBOAT CHANGING VELOCITY I—WORK & KINETIC ENERGY BAR CHART ANSW
a) The velocity of a tugboat increases from 2 m/s to 4 m/s in the same direction while a force is applied to the
tugboat for 20 seconds.
Fill in the missing bars for the work & kinetic energy bar chart for
this process.
Explain.
Initial
system
energy
KE
During
Wext
Final
system
energy
Bar chart key
KE
KE Kinetic energy
Wext Work done by
external forces
20 seconds later:
2 m/s
4 m/s
0
b) The velocity of a tugboat changes from 2 m/s to 4 m/s in the other direction while a force is applied to the tugboat
for 20 seconds.
Fill in the missing bars for the work & kinetic energy bar chart for this process.
Initial
system
energy
20 seconds later:
KE
During
Wext
Final
system
energy
Bar chart key
KE
KE Kinetic energy
4 m/s
2 m/s
Explain.
Wext Work done by
external forces
Answers (a) and (b) will have the same bar chart graphs. In both cases,
the speed of the tugboat doubles and so the kinetic energy quadruples.
In both cases, the same
net work will have been
20 seconds later:
done on the tugboat.
2 m/s
0
4 m/s
8) TUGBOAT CHANGING VELOCITY II—WORK & KINETIC ENERGY
BAR CHART
Initial
system
energy
a) The velocity of a tugboat changes from 2 m/s west to 4 m/s west while a
force is applied to the tugboat for 20 seconds.
KE
During
Wext
Final
system
energy
Bar chart key
KE
KE Kinetic energy
Wext Work done by
external forces
Draw a work & kinetic energy bar chart for this process.
Explain.
0
Answer: Since the speed has doubled, the final kinetic energy will be four
times larger than the initial kinetic energy. Positive work must have been
done on the tugboat.
b) The velocity of a tugboat changes from 4 m/s west to 2 m/s west while a
force is applied to the tugboat for 20 seconds.
Draw a work & kinetic energy bar chart for this process.
Initial
system
energy
KE
During
Wext
Final
system
energy
Bar chart key
KE
KE Kinetic energy
Explain.
Answer:
Since the
speed has
20 seconds later:
halved, the
final kinetic
4 m/s
2 m/s
energy must
be onequarter the initial kinetic energy. Negative work must have been done on
the tugboat.
Wext Work done by
external forces
0
9) BOX PULLED ON ROUGH SURFACE—WORK & KINETIC ENERGY BAR CHARTS ANSW
A 100 N box is initially at rest on a rough horizontal surface where the coefficient of static friction is 0.6 and the
coefficient of kinetic friction is 0.4. A student decides to
80 N
move the box by applying a horizontal force of 80 N to
100 N
the box to the right as shown. The box starts at rest at
point A.
5m
A
B
Fill in the work–kinetic energy and summary bar
charts below for the box as it moves between points A
and B. The chart at left below has columns for work done by four kinds of external force. In the summary chart at
right below you should include the net work done on the box as it moves from A to B.
Work-Energy Bar Chart
Initial
system
energy
KE
During
Wgrav
Wspring
Wfrict
Summary Chart
Initial
Final
system During system
energy
energy
Final
system
energy
Wapp
KE
KE
0
Wnet
KE
Bar chart key
KE Kinetic energy
Wgrav Work done by
gravitational forces
Wspring Work done by
spring forces
Wfrict Work done by
friction forces
0 Wapp Work done by
applied forces
Wnet Net work done by
all external forces
Use g = 10 m/s2
for simplicity
Explain.
Answer: The applied force has to do enough work to both increase the KE of the box and to compensate for the
negative work done by the frictional force. The applied force will do
(80 N)(5 m) = +400 J of work. The frictional force will do (40 N)(5 m) = 200 J of work. So the final kinetic
energy will be 200 J, which is the net work done.