Work and Energy Review
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Answers to Questions: All || #1-10 || #11-25 || #26-36 || #37-45
NOTE: The next 15 questions presume that the value of g is 10 m/s/s.
11. 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.
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a. less
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b. the same
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c. twice as much
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d. four times as much
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e. more than 4 times as much
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Answer: C
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.
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12. 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.
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a. 50
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b. more than 50
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c. less than 50
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Answer: A
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.
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13. 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?
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a. zero J
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b. more than zero J
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c. more information needed to determine
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Answer: A
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. A detailed discussion of a similar situation (the waiter and the tray of food) can be found at The Physics Classroom Tutorial |
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14. A 1000-kg car is moving at 40.0 km/hr when the driver slams on the brakes and skids to a stop (with locked brakes) over a distance of 20.0 meters. How far will the car skid with locked brakes if it is traveling at 120. km/hr?
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a. 20.0 m
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b. 60.0 m
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c. 90.0 m
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d. 120. m
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e. 180. m
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Answer: E
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*v2). For this reason, the skidding distance is directly proportional to the square of the speed. So if the speed 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). A detailed discussion of the distance-speed squared relationship can be found at The Physics Classroom Tutorial.
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15. 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.
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a. 10
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b. 500
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c. 510
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d. 5000
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e. more than 5000 .
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Answer: D
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.
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16. 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
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a. less than 100
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b. 100
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c. more than 100
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d. not enough information given
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Answer: A
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.
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17. 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.
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a. 2500
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b. 25000
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c. 250,000
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d. 2,500,000
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Answer: C
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.
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18. A 10-Newton object moves to the left at 1 m/s. Its kinetic energy is approximately ____ Joules.
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a. 0.5
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b. 1
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c. 10
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d. more than 10
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Answer: A
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.
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19. 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.
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a. less than 100
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b. 100
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c. 200
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d. more than 200
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e. none of these
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Answer: B
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 memorize the answer.
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20. An object at rest may have __________.
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a. speed
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b. velocity
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c. acceleration
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d. energy
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e. all of these
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Answer: D
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.
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21. 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.
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Answer: B
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
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22. 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.
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a. energy, work
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b. power, work
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c. work, energy
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d. work, power
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e. power, energy
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f. force, work
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g. power, force
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h. none of these
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Answer: D
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.
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23. Which requires more work: lifting a 50.0 kg crate a vertical distance of 2.0 meters or lifting a 25.0 kg crate a vertical distance of 4.0 meters?
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a. lifting the 50 kg crate
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b. lifting the 25 kg crate
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c. both require the same amount of work
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Answer: C
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.
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Lifting a 50 kg crate vertically 2 meters
W = (~500 N)*(2 m)*cos(0)
W = ~1000 N
(Note: The weight of a 50-kg object is approximately 500 N; it takes 500 N to lift the object up.)
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Lifting a 25 kg crate vertically 4 meters
W = (~250 N)*(4 m)*cos(0)
W = ~1000 N
(Note: The weight of a 25-kg object is approximately 250 N; it takes 250 N to lift the object up.)
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24. A 50.0 kg crate is lifted to a height of 2.0 meters in the same time as a 25.0 kg crate is lifted to a height of 4 meters. The rate at which energy is used (i.e., power) in raising the 50.0 kg crate is ____ as the rate at which energy is used to lift the 25.0 kg crate.
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a. twice as much
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b. half as much
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c. the same
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Answer: C
The power is the rate at which work is done (or energy is used). Power is found by dividing work by time. It requires the same amount of work to do these two jobs (see question #23) and the same amount of time. Thus, the power is the same for both tasks.
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25. Using 1000. J of work, a small object is lifted from the ground floor to the third floor of a tall building in 20.0 seconds. What power was required in this task?
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a. 20 W
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b. 50 W
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c. 100 W
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d. 1000 W
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e. 20000 W
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Answer: B
This is a relatively simple plug-and-chug into the equation P=W/t with W=1000. J and t=20.0 s.
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Navigate to:
Review Session Home - Topic Listing
Work and Energy - Home || Printable Version || Questions with Links
Answers to Questions: All || #1-10 || #11-25 || #26-36 || #37-45
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