# Chapter Test : Work, Energy And Power

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Questions pertain to the analysis of motion using relationships related to work and energy, mainly energy conservation and work-energy transfer principles. The following concepts are emphasized: work, positive work, negative work, energy, power, conservative (internal) forces, non-conservative (external) forces, potential energy, kinetic energy, mechanical energy, conservation of energy, work-energy theorem, pendulum, and incline planes.

• 1.

### Which of the following statements are true about work?

• A.

Work is a form of energy.

• B.

A Watt is the standard metric unit of work.

• C.

Work is a time-based quantity; it is dependent upon how fast a force displaces an object.

• D.

None of the above.

A. Work is a form of energy.
Explanation
Work is a form of energy, and in fact it has units of energy.

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• 2.

### Which of the following statements are true about power?

• A.

Powerful people or powerful machines are simply people or machines which always do a lot of work.

• B.

Power refers to how fast work is done upon an object.

• C.

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.

• D.

None of the above.

B. Power refers to how fast work is done upon an object.
Explanation
This is the definition of power.

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• 3.

### Which of the following statements are true about conservative and non-conservative forces?

• A.

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.

• 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.

None of the above.

C. The force of gravity and elastic (spring) force are both examples of a conservative forces.
Explanation
You must know this!

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• 4.

### Which of the following statements are true about kinetic energy?

• A.

If an object is on the ground, then it does not have any kinetic energy.

• B.

If an object is at rest, then it does not have any kinetic energy.

• C.

Kinetic energy is the form of mechanical energy which depends upon the position of an object.

• D.

None of the above.

B. If an object is at rest, then it does not have any kinetic energy.
Explanation
Kinetic energy depends upon speed. If there is no speed (the object is at rest), then there is no kinetic energy.

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• 5.

### Which of the following statements are true about potential energy?

• A.

Moving objects cannot have potential energy.

• B.

Both gravitational and elastic potential energy are dependent upon the mass of an object.

• C.

Potential energy is the energy stored in an object due to its position.

• D.

None of the above.

C. Potential energy is the energy stored in an object due to its position.
Explanation
This is the definition of potential energy.

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• 6.

### Which of the following statements are true about mechanical energy?

• 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.

None of the above.

A. The total amount of mechanical energy of an object is the sum of its potential energy and the kinetic energy.
Explanation
This is the definition of mechanical energy.

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• 7.

### 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. Presume that the value of g is ~10 m/s/s.

• A.

The same

• B.

Less

• C.

Twice as much

• D.

None of the above.

C. Twice as much
Explanation
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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• 8.

### An arrow is drawn back so that 50 Joules of potential energy is stored in the stretched bow and string. Whenreleased, the arrow will have a kinetic energy of ____ Joules. Presume that the value of g is ~10 m/s/s.

• A.

50

• B.

More than 50

• C.

Less than 50

• D.

None of the above.

A. 50
Explanation
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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• 9.

### 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? Presume that the value of g is ~10 m/s/s.

• A.

Zero J

• B.

More than zero J

• C.

• D.

None of the above.

A. Zero J
Explanation
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.

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• 10.

### 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? Presume that the value of g is ~10 m/s/s.

• A.

20.0 m

• B.

60.0 m

• C.

180. m

• D.

None of the above.

C. 180. m
Explanation
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).

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• 11.

### 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. Presume that the value of g is ~10 m/s/s.

• A.

5000

• B.

500

• C.

10

• D.

None of the above.

A. 5000
Explanation
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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• 12.

### 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. Presume that the value of g is ~10 m/s/s.

• A.

Less than 100

• B.

Not enough information given

• C.

More than 100

• D.

None of the above.

A. Less than 100
Explanation
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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• 13.

### 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. Presume that the value of g is ~10 m/s/s.

• A.

2500

• B.

25000

• C.

250,000

• D.

None of the above.

C. 250,000
Explanation
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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• 14.

### A 10-Newton object moves to the left at 1 m/s. Its kinetic energy is approximately ____ Joules. Presume that the value of g is ~10 m/s/s.

• A.

More than 10

• B.

0.5

• C.

10

• D.

None of the above.

B. 0.5
Explanation
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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• 15.

### 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. Presume that the value of g is ~10 m/s/s.

• A.

Less than 100

• B.

100

• C.

More than 200

• D.

None of the above.

B. 100
Explanation
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.

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• 16.

### An object at rest may have __________. Presume that the value of g is ~10 m/s/s.

• A.

Energy

• B.

Velocity

• C.

Acceleration

• D.

None of the above.

A. Energy
Explanation
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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• 17.

### 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. Presume that the value of g is ~10 m/s/s.

• A.

5

• B.

10

• C.

50

• D.

None of the above.

B. 10
Explanation
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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• 18.

### 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. Presume that the value of g is ~10 m/s/s.

• A.

Work, power

• B.

Force, work

• C.

Energy, work

• D.

None of the above.

A. Work, power
Explanation
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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• 19.

### 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? Presume that the value of g is ~10 m/s/s.

• A.

Lifting the 50 kg crate

• B.

Both require the same amount of work

• C.

Lifting the 25 kg crate

• D.

None of the above.

B. Both require the same amount of work
Explanation
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 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.)

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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• 20.

### 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. Presume that the value of g is ~10 m/s/s.

• A.

Twice as much

• B.

Half as much

• C.

The same

• D.

None of the above.

C. The same
Explanation
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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• 21.

### 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?  Presume that the value of g is ~10 m/s/s.

• A.

20 W

• B.

100 W

• C.

50 W

• D.

None of the above.

C. 50 W
Explanation
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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• 22.

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

• A.

~150 J

• B.

~250 J

• C.

~50 J

• D.

None of the above.

A. ~150 J
Explanation
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=24.5 N (it takes 24.5 N of force to lift a 2.5-kg object; that's the weight of the 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) = (24.5 N)*(6 m)*cos(0) = ~150 J (147 J)

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• 23.

### A student applies a force to a cart to pull it up an inclined plane at a constant speed during a physics 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.

• A.

12 J

• B.

13.2 J

• C.

23.5 J

• D.

None of the above.

B. 13.2 J
Explanation
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. (The angle theta represents the angle betwee the force and the displacement vector; since the force is applied parallel to the incline, the angle is zero.) 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 at constant speed changes the potential energy of the cart. The work done equals the potential energy change. Thus,

W=Delta PE = m*g*(delta h) = (3.00 kg)*(9.8 m/s/s)*(0.45 m) = 13.2 J

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• 24.

### Eddy, whose mass is 65.0-kg, climbs up the 1.60-meter high stairs in 1.20 s. Approximate Eddy's power rating.

• A.

P = 849 Watts

• B.

P = 749 Watts

• C.

P = 659 Watts

• D.

None of the above.

A. P = 849 Watts
Explanation
Eddy's power is found by dividing the work which he does by the time in which he does it. The work done in elevating his 65.0-kg mass up the stairs is determined using the equation
W = F*d*cos(Theta)
where F = m*g = 637 N (the weight of the 65.0 kg object), d =1.60 m and Theta = 0 degrees (the angle between the upward force and the upward displacement). Solving for W yields 1019.2 Joules. Now divide the work by the time to determine the power:
P = W/t = (1019.2 J)/(1.20 s) = 849 Watts

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• 25.

### A 65.8-kg skier accelerates down an icy hill from an original height of 521 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.  (b) 1.40*105 J of energy are lost due to external forces.

• A.

(a) v = 101 m/s; (b) v = 77.2 m/s

• B.

(a) v = 77.2 m/s; (b) v = 101 m/s

• C.

(a) v = 151 m/s; (b) v = 87.2 m/s

• D.

None of the above.

A. (a) v = 101 m/s; (b) v = 77.2 m/s
Explanation
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=65.8 kg, h=521 m, g=9.8 m/s/s. Substituting and solving for vf yields 101 m/s.

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

m*g*h - 140000 J = 0.5*m*vf2
Now substituting and solving for vf yields 77.2 m/s.

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• 26.

### Use the work-energy theorem to determine the force required to stop a 988-kg car moving at a speed of 21.2 m/s if there is a distance of 45.7 m in which to stop it.

• A.

F = 14.86*103 N

• B.

F = 4.86*103 N

• C.

F = 5.60*103 N

• D.

None of the above.

B. F = 4.86*103 N
Explanation
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=988 kg, vi=21.2 m/s, d=45.7 m, and Theta = 180 degrees. Substituting and solving for F yields 4.86*103 N.

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• 27.

### 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 is 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.

• A.

2.56 m/s

• B.

3.56 m/s

• C.

12.56 m/s

• D.

None of the above.

A. 2.56 m/s
Explanation
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.

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• 28.

### 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.

• A.

163 N

• B.

514 N

• C.

251 N

• D.

None of the above.

B. 514 N
Explanation
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 â€¢ m â€¢ vi2) from the final value (0 J) .

Change in KE = KEf - KEi = 0 J - 0.5 â€¢ (0.163 kg) â€¢ (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 â€¢ d â€¢ cos(theta) = -129 J

F â€¢ (0.251 m) â€¢ cos(180 deg) = -129 J

F = 514 N

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• 29.

### 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?

• A.

629 m

• B.

121 m

• C.

116 m

• D.

None of the above.

C. 116 m
Explanation
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
Ffrict = 72.3 N

On Level Surface
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

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

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

44541 J = 384 x

116 m = x

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• 30.

### A 29.1-kg sledder is traveling along a level area with a speed of 8.96 m/s when she approaches a gentle incline which makes an angle of 12.5 degrees with the horizontal. If the coefficient of friction between the sled and the incline is 0.109, then what will be her speed at the bottom of the inclined plane, located 8.21 m above the top of the incline.

• A.

12.7 m/s

• B.

291 m/s

• C.

100 m/s

• D.

None of the above.

A. 12.7 m/s
Explanation
This problem is similar to the above. During the entire descent down the hill, gravity is doing work on the sledder and friction is doing negative work on the sledder. Friction is a non-conservative force and will alter the total mechanical energy of the sledder. The equation to be used is

KEi + PEi + Wnc = KEf + PEf

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• 31.

### 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.

• A.

36.7 m/s

• B.

46.1 m/s

• C.

461 m/s

• D.

None of the above.

B. 46.1 m/s
Explanation
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

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• 32.

### Claire deAisles has just completed her shopping at the grocery 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.

• A.

927 N

• B.

149 N

• C.

218 N

• D.

None of the above.

C. 218 N
Explanation
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:

mâ€¢gâ€¢hi + Fâ€¢dâ€¢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)â€¢(9.8 m/s/s)â€¢(2.38 m) + Fâ€¢(0.0316 m)â€¢cos(180 deg) = 0 J
6.89 J - (0.0316m ) â€¢ F = 0 J

6.89 J = (0.0316m ) â€¢ F

F = 218 N

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• 33.

### 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?

• A.

0.119 m

• B.

0.170 m

• C.

0.149 m

• D.

None of the above.

B. 0.170 m
Explanation
The force of friction is related to the normal force (= mâ€¢g) and the coefficient of friction (0.728). The Ffrict is

Ffrict = muâ€¢Fnorm = (0.728) â€¢ (1.49 kg) â€¢ (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 â€¢ d â€¢ cos(theta) = (1.27 N) â€¢ (1.42 m) â€¢ 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 â€¢ d â€¢ cos(theta) = -1.80 J

(10.6 N) â€¢ (d) â€¢ cos(180 deg) = -1.80 J

- (10.6 N) â€¢ (d) = -1.80 J

d = 0.170 m

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• 34.

### 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)?

• A.

132 m

• B.

492 m

• C.

332 m

• D.

None of the above.

A. 132 m
Explanation
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
PEi + Wnc = 0
PEi + Wincline + Wlevel = 0

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

45102 J - 11307 J - 257 x = 0 J

33795 J = 257 x

132 m = x

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• 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.

• A.

5243 N

• B.

5423 N

• C.

2423 N

• D.

None of the above.

B. 5423 N
Explanation
mâ€¢gâ€¢hi + Fâ€¢dâ€¢cos(theta) = 0 J
(62.4 kg) â€¢ (9.8 m/s/s) â€¢ (7.91 m) + F â€¢ (0.892 m) â€¢ cos(180 deg) = 0
4837 J - (0.892 m) â€¢ F = 0

4837 J = (0.892 m) â€¢ F

F = 5423 N

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• 36.

### A car having a mass of 500 kg is initially traveling with a speed of 80 km/hr. It slows down at a constant rate, coming to a stop in a distance of 50 m. What is the change in the car's kinetic energy over the 50 m distance it travels while coming to a stop?

• A.

1.2x10^5 J

• B.

-1.2x10^5 J

• C.

-11.2x10^5 J

• D.

None of the above.

B. -1.2x10^5 J
Explanation
Initially, the car has a speed of 80 km/hr so has a KE of 1/2 Mv2; we need to be sure to convert the speed into SI units. After 50 m, the car is at rest, so its KE is zero.

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• 37.

### A car having a mass of 500 kg is initially traveling with a speed of 80 km/hr. It slows down at a constant rate, coming to a stop in a distance of 50 m. What is the net force on the car while it's coming to a stop?

• A.

-12.4x10^3 N

• B.

2.4x10^3 N

• C.

-2.4x10^3 N

• D.

None of the above.

C. -2.4x10^3 N
Explanation
The total work done on the car (the work done by the net force) equals the change in KE.
total work done on car = Fnetd = Fnet x 50 m
total work done on car = change in KE
-> Fnet = -1.2 x 105J / 50 m = -2.4 x 103 N

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• 38.

### A car having a mass of 500 kg is initially traveling with a speed of 80 km/hr. It slows down at a constant rate, coming to a stop in a distance of 50 m. Where is the force applied?

• A.

The force on the car causing it to stop is applied at the only point of contact the car has -- the ground! The ground pushes on the car opposite to the car's motion (that's the meaning of the minus sign above), causing it to stop. You can tell this, because a car's tires get hot from the friction forces between them and the road.

• B.

Kinetic energy is the form of mechanical energy which depends upon the position of an object.

• C.

Faster moving objects always have a greater kinetic energy.

• D.

None of the above.

A. The force on the car causing it to stop is applied at the only point of contact the car has -- the ground! The ground pushes on the car opposite to the car's motion (that's the meaning of the minus sign above), causing it to stop. You can tell this, because a car's tires get hot from the friction forces between them and the road.
Explanation
The force on the car causing it to stop is applied at the only point of contact the car has -- the ground! The ground pushes on the car opposite to the car's motion (that's the meaning of the minus sign above), causing it to stop. You can tell this, because a car's tires get hot from the friction forces between them and the road.

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• 39.

### A car having a mass of 500 kg is initially traveling with a speed of 80 km/hr. It slows down at a constant rate, coming to a stop in a distance of 50 m. If the car slows to a stop on level ground, is the work done on it recoverable?

• A.

The KE of the cars motion is turned into heat energy (the car's tires, the brakes and the road get hot) so the work done on the car is not recoverable.

• B.

Faster moving objects always have a greater kinetic energy.

• C.

Kinetic energy is the form of mechanical energy which depends upon the position of an object.

• D.

None of the above.

A. The KE of the cars motion is turned into heat energy (the car's tires, the brakes and the road get hot) so the work done on the car is not recoverable.
Explanation
The correct answer explains that when the car slows down, the kinetic energy of its motion is converted into heat energy. This energy is dissipated through the car's tires, brakes, and the road, making it impossible to recover the work done on the car. This is because energy is lost in the form of heat, and it cannot be converted back into useful work. Therefore, the work done on the car is not recoverable.

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• 40.

### A 1 kg ball is dropped from the top of a cliff and falls with a constant acceleration due to gravity (9.8 m/s2). Assume that effects of air resistance can be ignored.  By how much has the ball's gravitational potential energy changed after it has fallen by 10 m?

• A.

-98 J

• B.

98 J

• C.

120 J

• D.

None of the above.

A. -98 J
Explanation
The ball's potential energy (PE) gets smaller by an amount Mgh, where h is the height through which it falls (10 m). This energy change results from the work done on the ball by gravity. Numerically,
change in PE = -Mgh = -(1.0 kg)(9.8 m/s2)(10 m) = -98 J

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• 41.

### A 1 kg ball is dropped from the top of a cliff and falls with a constant acceleration due to gravity (9.8 m/s2). Assume that effects of air resistance can be ignored. How fast is the ball going after it has fallen by 10 m?

• A.

45 m/s

• B.

14 m/s

• C.

87 m/s

• D.

None of the above.

B. 14 m/s
Explanation
We can find the speed in one of two ways. The first way is to realize the ball has a constant acceleration (9.8 m/s2 downward), and we can then find the time it takes for the ball to fall. This is harder than the second way. Since the work done by gravity is recoverable we can say that the PE lost equals the kinetic energy (KE) gained by the ball. Initially, the ball's KE is zero.

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• 42.

### A 1 kg ball is dropped from the top of a cliff and falls with a constant acceleration due to gravity (9.8 m/s2). Assume that effects of air resistance can be ignored. What is the force (if any) that does work on the ball?

• A.

It's the force of gravity that does work on the ball.

• B.

F work is done on an object by a non-conservative force, then the object will either gain or lose kinetic energy.

• C.

More massive objects always have a greater kinetic energy.

• D.

None of the above.

A. It's the force of gravity that does work on the ball.
Explanation
The force of gravity does work on the ball because as the ball falls, gravity exerts a force on it in the direction of motion. This force causes the ball to accelerate and gain kinetic energy. Since work is defined as the transfer of energy, the force of gravity is doing work on the ball.

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• 43.

### A 1 kg ball is dropped from the top of a cliff and falls with a constant acceleration due to gravity (9.8 m/s2). Assume that effects of air resistance can be ignored. How much work has been done on the ball?

• A.

A falling object always gains kinetic energy as it falls.

• B.

An object can never have a negative kinetic energy.

• C.

The work done on the ball equals the energy changed from PE to KE, or 98 J.

• D.

None of the above.

C. The work done on the ball equals the energy changed from PE to KE, or 98 J.
Explanation
When the ball is dropped from the cliff, it falls under the influence of gravity. As it falls, it gains kinetic energy, which is equal to the potential energy it loses. This is because the work done on the ball is equal to the change in energy from potential energy (PE) to kinetic energy (KE). Since the ball has a mass of 1 kg and the acceleration due to gravity is 9.8 m/s^2, the work done on the ball can be calculated as the product of the mass, acceleration due to gravity, and the height of the cliff. Therefore, the work done on the ball is 1 kg * 9.8 m/s^2 * 10 m = 98 J.

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• 44.

### A 1 kg ball is dropped from the top of a cliff and falls with a constant acceleration due to gravity (9.8 m/s2). Assume that effects of air resistance can be ignored. Is the work done on the ball recoverable?

• A.

An object can never have a negative kinetic energy.

• B.

Technically no, the work done is not increasing the potential energy, it is decreasing it! We haven't converted energy into a form that we can tap into again.

• C.

An object has a kinetic energy of 40 J. If its mass were twice as much, then its kinetic energy would be 80 J.

• D.

None of the above.

B. Technically no, the work done is not increasing the potential energy, it is decreasing it! We haven't converted energy into a form that we can tap into again.
Explanation
The work done on the ball is not recoverable because it is not increasing the potential energy of the ball. Instead, it is decreasing the potential energy. This means that we have not converted the energy into a form that we can tap into again.

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• 45.

### A car having a mass of 500 kg is initially at rest. A constant 1,000 N net force acts on the car over a distance of 50 m, causing the car to speed up. After it travels 50 m, the car moves with constant velocity. What is the total work done on the car over the 50 m distance it travels while speeding up?

• A.

25x10^4 J

• B.

5x10^4 J

• C.

15x10^4 J

• D.

None of the above.

B. 5x10^4 J
Explanation
The total work done on the car is the work done by the net force.
total work = Fnetd = (1,000 N)(50 m) = 5x10^4J

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• 46.

### A car having a mass of 500 kg is initially at rest. A constant 1,000 N net force acts on the car over a distance of 50 m, causing the car to speed up. After it travels 50 m, the car moves with constant velocity. How fast is the car moving after 50 m?

• A.

24 m/s

• B.

14 m/s

• C.

34 m/s

• D.

None of the above.

B. 14 m/s
Explanation
The total work changes the car's kinetic energy (KE). Initially, the car is at rest, so its KE is zero. After traveling a distance of 50 m, the car moves with some unknown speed v,

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• 47.

### A car having a mass of 500 kg is initially at rest. A constant 1,000 N net force acts on the car over a distance of 50 m, causing the car to speed up. After it travels 50 m, the car moves with constant velocity. What is the net force on the car while its moving with constant velocity? d.) What is the total work done on the car while its moving with constant velocity?

• A.

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.

• B.

When the car moves with constant velocity, its acceleration is zero. By Newton's second law, this means that the net force is zero.

• C.

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.

• D.

None of the above.

B. When the car moves with constant velocity, its acceleration is zero. By Newton's second law, this means that the net force is zero.
Explanation
When an object is moving with constant velocity, it means that its acceleration is zero. According to Newton's second law of motion, the net force acting on an object is equal to the product of its mass and acceleration. Since the acceleration is zero, the net force on the car while it is moving with constant velocity is also zero. This means that there is no additional force acting on the car to change its velocity.

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• 48.

### A car having a mass of 500 kg is initially at rest. A constant 1,000 N net force acts on the car over a distance of 50 m, causing the car to speed up. After it travels 50 m, the car moves with constant velocity. What is the total work done on the car while its moving with constant velocity?

• A.

The total work is the work done by the net force. After 50 m, the car moves with constant velocity so the net force is zero. This means that the total work done on the car after 50 m is zero. Another way to see this is that the car's speed, hence its KE, isn't changing. This means the total work is zero.

• B.

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

• C.

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.

• D.

None of the above.

A. The total work is the work done by the net force. After 50 m, the car moves with constant velocity so the net force is zero. This means that the total work done on the car after 50 m is zero. Another way to see this is that the car's speed, hence its KE, isn't changing. This means the total work is zero.
Explanation
The total work done on the car while it is moving with constant velocity is zero. This is because when the car moves with constant velocity, the net force acting on it is zero. Therefore, no work is being done on the car by the net force. Additionally, since the car's speed and kinetic energy are not changing, there is no change in the work done on the car. Hence, the total work done on the car after it travels 50 m is zero.

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• 49.

### Calculate the kinetic energy (KE) of a 1500-kg automobile with a speed of 30 m/s.

• A.

12.8 x 10 ^ 5 J

• B.

6.8 x 10 ^ 5 J

• C.

9.8 x 10 ^ 5 J

• D.

None of the above.

B. 6.8 x 10 ^ 5 J
Explanation
KE = 1/2 Mv2 = 1/2 (1500 kg)(30 m/s)2 = 6.8 x 10 ^5 J

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• 50.

### Calculate the kinetic energy (KE) of a 1500-kg automobile with a speed of 30 m/s. If it accelerates to this speed in 20 s, what average power has been developed?

• A.

3.4 x 10 ^ 4 W

• B.

13.4 x 10 ^ 4 W

• C.

6.4 x 10 ^ 4 W

• D.

None of the above.

A. 3.4 x 10 ^ 4 W
Explanation
KE = 1/2 Mv2 = 1/2 (1500 kg)(30 m/s)2 = 6.8 x 10 ^ 5 J
The total work done equals 6.8 x 10 ^ 5 J; i.e., this is the change in kinetic energy. The average power is the total work divided by the time interval, or

average power = 6.8 x 105 J / 20 s = 3.4 x 10 ^ 4 W

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