# Units 4-5 Review

Momentum and Energy

 Navigate to Answers for: [ Questions #1-#14 | Questions #15-#28 | Questions #29-#42 | Questions #43-#56 | Questions #57-#66 ] [ #1 | #2 | #3 | #4 | #5 | #6 | #7 | #8 | #9 | #10 | #11 | #12 | #13 | #14 ]

1. An object is raised above the ground gaining a certain amount of potential energy. If the same object is raised twice as high it gains __________.

1. four times as much potential energy
2. twice as much potential energy
3. neither of these

The potential energy of an object is directly proportional to the height of the object above the ground (or some other arbitrary "zero-level") in accordance with the equation

PE = m*g*h

If the h in the equation is doubled (the object is raised twice as high), then the PE is doubled as well.

Potential Energy (12 seconds)

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2. When an object is lifted 10 meters, it gains a certain amount of potential energy. If the same object is lifted 20 meters, its potential energy is _____.
 a. less b. the same c. twice as much d. four times as much e. more than 4 time as much

The potential energy of an object is directly proportional to the height of the object above the ground (or some other arbitrary "zero-level") in accordance with the equation

PE = m*g*h

If the h in the equation is doubled (say from 10 m to 20 m), then the PE is doubled as well.

Potential Energy (12 seconds)

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3. A 1000 kg car and a 2000 kg car are hoisted the same distance at constant speed in a gas station. Raising the more massive car requires ____________.
 a. less work b. as much work c. twice as much work. d. four times as much work e. more than 4 times as much work

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 2000-kg car weight 2X as much as the 1000-kg car, it would require twice as much work to lift it the same distance.

Definition and Mathematics of Work (11 seconds) | Potential Energy (12 seconds) | Analysis of Situations Involving External Forces (21 seconds)

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4. An object that has kinetic energy must be ____________.
 a. moving b. falling c. at an elevated position d. at rest e. none of these

Pay attention to the keywords "must be." Kinetic energy is the energy of motion and an object must be moving if it has kinetic energy. The object could be falling and could be at an elevated position; the object must not be at rest if it has kinetic energy.

Kinetic Energy (4 seconds)

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5. An object that has potential energy has this energy because of its _____________.
 a. speed b. acceleration c. momentum d. position e. none of these

Potential energy is the energy of position and any object which has potential energy owes this PE to the fact that it has some given position other than the so-called "zero-level position."

Potential Energy (12 seconds)

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6. An arrow is drawn so that it has 40 J of potential energy. When fired horizontally, the arrow will have a kinetic energy of ________.
 a. less than 40 J b. more than 40 J c. 40 J

A drawn arrow has 40 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 40 J of kinetic energy upon being fired. Of course, this assumes no energy is lost to air resistance, friction or any other external forces and that the arrow is shot horizontally.

Mechanical Energy (15 seconds)

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7. A 2 kg mass is held 4 m above the ground. What is the approximate potential energy of the mass with respect to the ground?

 a. 20 J b. 40 J c. 60 J d. 80 J e. none of these

The potential energy of an object is found from the equation

PE = m*g*h

where m is mass, h is height, and g=10 m/s/s (approx.). Plugging and chugging into this equation yields PE=(2 kg)*(10 m/s/s)*(4 m) = 80 J.

Potential Energy (12 seconds)

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8. A 2 kg mass has 40 J of potential energy with respect to the ground. Approximately how far is it located above the ground?
 a. 1 m b. 2 m c. 3 m d. 4 m e. none of these

The potential energy of an object is related to its height by the equation

PE = m*g*h

where m is mass, h is height, and g=10 m/s/s (approx.). Plugging and chugging into this equation yields 40 J = (2 kg)*(10 m/s/s)*(h); rearranging and solving for h yields an answer of 2 m.

Potential Energy (12 seconds)

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9. A ball is projected into the air with 100 J of kinetic energy which is transformed to gravitational potential energy at the top of its trajectory. When it returns to its original level after encountering air resistance, its kinetic energy is ____________.
 a. less than 100 J b. 100 J c. more than 100 J d. not enough information given

During any given motion, if external forces do work upon the object, then the total mechanical energy will be changed. If external forces do negative work (i.e., F*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.

Mechanical Energy Conservation (13 seconds)

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10. A woman lifts a box from the floor. She then carries with constant speed to the other side of the room, where she puts the box down. How much work does she do on the box while walking across the floor at constant speed?
 a. zero J b. more than zero J c. more information needed to determine

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 displacment of the object, and Theta is the angle between the force and the displacement. In this specific situation, the woman is applying an upward force on the box (she 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 woman 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.

Definition and Mathematics of Work (10 seconds)

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11. A car moving at 50 km/hr skids 20 m with locked brakes. How far will the car skid with locked brakes if it is traveling at 150 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 50 km/hr to 150 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.

Analysis of Situations Involving External Forces (21 seconds)

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12. Which has greater kinetic energy, a car traveling at 30 km/hr or a half-as-massive car traveling at 60 km/hr?
 a. the 30 km/hr car b. the 60 km/hr car c. both have the same kinetic energy

This problem is complicated by the fact that no mass is given for the two cars; only the ratio of mass is known. The complication can be resolved in one of two ways: 1) make up a mass for each car - such as 10 kg and 5 kg, or 2) assign m as the mass of one of the cars and (1/2)m as the mass of the second car. Then use the kinetic equation

KE = 0.5*m*v2

and plug and chug. The mass can then be determined for each car and compared. Using the second method yields the following results for the two cars:

 KE for the 30 km/hr car: KE = 0.5*m*(30 km/hr)2 KE = 0.5*m*(900 km2/hr2) KE = 450*m km2/hr2 KE for the 60 km/hr car: KE = 0.5*(0.5m)*(60 km/hr)2 KE = 0.5*(0.5m)*(3600 km2/hr2) KE = 900*m km2/hr2

Kinetic Energy (4 seconds)

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13. A diver who weighs 500 N steps off a diving board that is 10 m above the water. The diver hits the water with kinetic energy of ___________.
 a. 10 J b. 500 J c. 510 J d. 5000 J e. more than 5000 J.

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.

Kinetic Energy (4 seconds) | Mechanical Energy Conservation (13 seconds)

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14. A 2500 N pile driver ram falls 10 m and drives a post 0.1 m into the ground. The average impact force on the ram is _________.
 a. 2500 N b. 25000 N c. 250,000 N d. 2,500,000 N

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; external work has been done by an applied force upon the pile driver. 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 + Wext = 0

PEi = - Wext

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 m for h; 0.1 m for the displacement of the driver as caused by the upward applied force exerted by the ram; and 90 degrees for Theta (the angle between the applied force and the displacement of the ram) will yield the answer of 250000 N for F.

Analysis of Situations Involving External Forces (21 seconds)

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