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Unit 5 Quiz#2


Equations:


Use the diagram at the right to answer questions #1 and #2.

1. As the pendulum bob swings from position A to position B, its total mechanical energy (neglecting friction and air resistance)

a. decreases.

b. increases.

c. remains the same.

ANSWER: The answer is C. Since there are no external forces doing work (recall the "Energy of a Pendulum" Lab), the total mechancial energy is conserved - i.e. , everywhere the same.


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Mechanical Energy (15 seconds)

 

2. As the pendulum bob swings from position A to position B, its

a. speed increases

b. potential energy decreases

c. mass decreases

d. kinetic energy decreases

e. kinetic energy remains constant

ANSWER: The answer is D. Since the object increases its PE (due to a height increase from A to B), there is a decrease in kinetic energy due to the slowing down of the pendulum bob (recall the "Energy of a Pendulum" Lab).


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Kinetic Energy (4 seconds)

Analysis of Situations in Which Mechanical Energy is Conserved (13 seconds)

 

3. Person X is able to lift a 50 kg barbell to a height of 2.0 m in 1.0 s. If person Y takes 1.5 s to do the same thing, then, in this lift

a. X does more work than Y.

b. Y does more work than X.

c. X develops more power than Y.

d. Y develops more power than X.

e. X and Y develop the same power.

ANSWER: The answer is C. The two people apply the same force to lift the same mass the same distance; thus, they do the same amount of work. Yet the person who does the same work in the least time is the most power-full.


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Work (10 seconds)

Power (13 seconds)

 

 

4. The kilowatt * hour is a unit of

a. power

b. force

c. velocity

d. work and energy

 

ANSWER: The answer is D. A kilowatt (similar to a watt) is a unit of power and the hour is a unit of time. Thus, a kilowatt*hour is a unit of power*time. From the equation list above, you ought to recognize that power*time is equal to work. Therefore, the kilowatt*hour is a unit of work or energy. This is an example of a unit analysis question which will appear periodically on future quizzes.


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Work (10 seconds)

Power (13 seconds)

 

Use the following diagram to answer questions #5 - #7. Neglect the effect of friction and air resistance.

5. As the object moves from point A to point D across the frictionless surface, the sum of its gravitational potential and kinetic energies

a. decreases, only.

b. decreases and then increases.

c. increases and then decreases.

d. remains the same.

ANSWER: The answer is D. The total mechanical energy (i.e., the sum of the kinetic and potential energies) is everywhere the same whenever there are no external forces (such as friction or air resistance) doing work.


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Work and Energy Practice Questions (10 seconds)

 

 

6. The object will have a minimum gravitational potential energy at point

a. A.

b. B.

c. C.

d. D.

E. E.

ANSWER: The answer is B. Gravitaional potential energy depends upon height (PE=m*g*h). The PE is a minimum when the height is a minimum. Position B is the lowest position in the diagram.


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Work and Energy Practice Questions (10 seconds)

 

 

7. The object's kinetic energy at point C is less than its kinetic energy at point

a. A only.

b. A, D, and E.

c. B only.

d. D and E.

ANSWER: The answer is C. Since the total mechanical energy is conserved, kinetic energy (and thus, speed) will be greatest when the potential energy is smallest. Point B is the only point that is lower than point C. The reasoning would follow that point C is the point with the smallest PE, the greatest KE, and the greatest speed. Therefore, the object will have less kinetic energy at point C than at point B (only).


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Work and Energy Practice Questions (10 seconds)

 

 

8. A person prevents a 7500 N car with its brakes released from rolling down a hill by pushing on the car with a force of 250 N. How much work does the person do to prevent the car from rolling?

a. 0 J

b. 5000 J

c. 150 000 J

d. 300 000 J

ANSWER: The answer is A. Since the force exerted by the person does not cause the car to be displaced, the person does no work upon the car. This question is similar to a unit packet question on p. 1 of the unit - "A physics teacher exerts a force on a wall for several hours and becomes exhausted; does the phyusics teacher do any work?".


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Work (10 seconds)

 

9. During a certain time interval, a 20-N object free-falls 10 meters. The object gains _____ Joules of kinetic energy during this interval.

a. 10

b. 20

c. 200

d. 2000

ANSWER: The answer is C. Energy is conserved in free-fall situations (no external forces doing work). Thus, whatever potential energy is lost is transformed into kinetic energy (refer to the "Energy Conservation on an Incline" Lab). The object loses 200 J of potential energy (PE loss = m * g * h loss = 20 kg * 10 m/s/s * 1 m = 200 J). Observe that a confusion of mass (20 kg) and weight (200 N) will inevitably lead to the wrong answer.


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Analysis of Situations in Which Mechanical Energy is Conserved (13 seconds)

 

 

10. An object which weighs 10 N is dropped from rest from a height of 4 meters above the ground. When it has free-fallen 1 meter its total mechanical energy with respect to the ground is

a. 2.5 J

b. 10 J

c. 30 J

d. 40 J

ANSWER: The answer is D. Energy is conserved in free-fall situations (no external forces doing work). Thus, the total mechanical energy initially is everywhere the same; whatever TME it has initially, it will maintain throughout the course of its motion. The object begins with 40 J of potential energy (PE = m * g * h = 1 kg * 10 m/s/s * 4 m = 40 J) and no kinetic energy. The total mechanical energy (KE + PE) is 40 J. Observe that a confusion of mass (1 kg) and weight (10 N) will inevitably lead to the wrong answer.


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Potential Energy (12 seconds)

Analysis of Situations in Which Mechanical Energy is Conserved (13 seconds)

 

 

 

PROBLEM-SOLVING: Please show all your work in an organized manner wherever indicated by PSYW.

11. The distance required for a car to brake to a stop is dependent upon its velocity before braking. Illustrate your understanding of the relationship between stopping distance and velocity by filling in the blanks in the following table.

ANSWER: As found in the Stopping Distance vs. Velocity Lab, the distance is proportional to the square of the velocity. Thus, as the velocity is increased by a factor of 2 (from 10 m/s to 20 m/s), the stopping distance is increased by a factor of 2^2 or 4 (from 8 m to 32 m). As the velocity is increased by a factor of 3 (from 10 m/s to 30 m/s), the stopping distance is increased by a factor of 3^2 or 9 (from 8 m to 72 m).

 


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Analysis of Situations Involving External Forces (21 seconds)

 

 

12. A Hot Wheels car traveling with a speed of 4.0 m/s is coasting along an elevated section of track and towards an incline (see diagram). Determine the height to which the car will rise on the opposite incline. Assume no friction or air resistance. PSYW

ANSWER: This is problem 2b from the unit packet handout titled "Energy Concepts." Review the solution as worked out in class. The answer is 2.3 meters.


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Analysis of Situations in Which Mechanical Energy is Conserved (13 seconds)

 

13. Near the end of the Shock Wave roller Coaster ride at Great America, a braking system applies a large force to bring the 5000-kg train of cars from a speed of approximately 20 m/s to a speed of 5 m/s over a distance of 20 meters (see diagram). Use the work-energy theorem to determine the magnitude of the force. PSYW

ANSWER: If you review your packet handout, you will see a nearly identical problem. The only difference between the problem done in class and this problem is that the unknown here is the force (in class, the unknown was the distance). The work-energy theorem is used

KEi + PEi + Wext = KEf + PEf

0.5 * 5000 * 20^2 + 0 + F * 20 * cos(180 deg) = 0.5 * 5000 * 5^2 + 0

1 000 000 J - 20 * F = 62 500 J

-20 * F = -937 500 J

F = 46 875 N

 


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Analysis of Situations Involving External Forces (21 seconds)

 

 

 

Li Ping Phar, the esteemed Chinese ski jumper, starts at rest on top of a 100-meter hill, skis down the 45° incline and makes a world-record setting jump. The heights at various locations along her path are shown below. Use the diagram to answer questions #14-#16. Assume that friction and air resistance has a negligible effect upon Li's motion and assume that Li never uses her poles for propulsion.

14. Carefully construct energy-bar charts for Li Ping Phar's motion for the five indicated positions (A through E).

ANSWER:


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Work and Energy Practice Questions (10 seconds)

 

 

15. Li Ping Phar has a mass of 50 kg. Determine Li Ping Phar's speed at position D. PSYW

ANSWER: Use the work-energy theorem to determine the speed at position D. Choosing position A to be the initial condition and position D to be the final position, the solution takes the following form

KEi + PEi + Wext = KEf + PEf

0 + 50 * 10 * 100 + 0 = 0.5 * 50 * v^2 + 50 * 10 * h

Cancel mass (50) from each term or merely work through the algebra as given to reduce to

50 000 = 25 * v^2 + 30 000

20 000 = 25 * v^2

800 = v^2

28.3 m/s = v


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Work and Energy Practice Questions (10 seconds)

 

 

16. The Swiss ski jumper, Eaton Alottafude, has a mass of 80 kg. Would Eaton's greater mass create an advantage, disadvantage, or make no difference to his success? Explain clearly and logically.

ANSWER: The answer is NO! In situations when there are no external forces acting upon an object, the energy is conserved and the mass makes no difference on the outcomes of motion parameters. Similar to free-fall situations, the work-energy theorem predicts that the final speeds of objects are independent of their mass in situations where friction and air resistance can be ignored. Since mass cancels from the equation, the greater mass of Eaton Allotafude makes no difference. If still uncertain, merely substitute 80 kg into the calculations above and see that the speed is uneffected by the greater mass.

 


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Work and Energy Practice Questions (10 seconds)

 

 

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Special thanks to lab assistant Carl Bobis for his assistance in converting quiz to HTML format.

 

© Tom Henderson, 1996-1998


Glenbrook South High School.

Last updated on 9/29/98.