# Units 5 Review

Work, Energy and Power

 Units 5 Review - Questions Only Navigate to Answers for:  [ #32 | #33 | #34 | #35 | #36 | #37 | #38 | #39 | #40 ]

32. Calculate the work required lift a 2.5-kg object a height of 6.0 meters. PSYW

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=25 N (it takes approx. 25 N of force to lift a 2.5-kg 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) = (25 N)*(6 m)*cos(0) = 150 J

Calculating the Amount of Work Done by Forces (10 seconds)

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33. In the It's All Uphill Lab, a force of 21.2 N is applied parallel to the incline to lift a 3.0-kg loaded cart to a height of 0.45 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. PSYW

There are two methods of solving this problem. The first method involves using the equation

W = F*d*cos(Theta)

where F=21.2 N, d=0.636 m, and Theta=0 degrees. Substituting and solving yields

W = F*d*cos(Theta) = (21.2 N)*(0.636 m)*cos(0) = 13.5 J.

The second method is to recognize that the work done in pulling the cart along the incline changes the potential energy of the cart. The work done equals the potential energy change. Thus,

W=Delta PE = m*g*(delta h) = (3 kg)*(10 m/s/s)*(0.45 m) = 13.5 J

Calculating the Amount of Work Done by Forces (10 seconds) | Analysis of Situations Involving External Forces (21 seconds)

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34. An 800-kg car skids to a stop across a horizontal surface over a distance of 45 m. The average force acting upon the car is 7000N, then determine

1. the work done upon the car.
2. the initial kinetic energy of the car.
3. the acceleration of the car.
4. the initial velocity of the car.

1. W = -315000 J
2. KEi = +315000 J
3. a = -8.75 m/s/s
4. vi = 28.1 m/s

a. The work done upon the car can be found using the equation

W = F*d*cos(Theta)

where F=7000 N, d=45 m, and Theta=180 degrees (the force is in the opposite dir'n as the displacement). Substituting and solving yields -315000 J.

b. The initial kinetic energy can be found using the work-energy theorem. The eq'n reduces to

KEi + Wext = 0

(PEi and PEf = 0 J since the car is on the ground; and KEf = 0 J since the car is finally stopped). Rearrange the equation and it takes the form KEi = -Wext . So KEi = +315000 J.

c. The acceleration of the car can be found using Newton's second law of motion: Fnet = m*a

The friction force is the net force (since the up and down forces balance) and the mass is 800 kg. Substituting and solving yields a = -8.75 m/s/s.

d. The initial velocity of the car can be found using the KE equation: KE = 0.5*m*v2 where m=800 kg and KEi=315000 J. Substituting and solving for velocity (v) yields v = 28.1 m/s. (A kinematic equation could be also used to find the initial velocity.)

Analysis of Situations Involving External Forces (21 seconds)

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35. A 50-kg hiker ascends a 40-meter high hill at a constant speed of 1.2 m/s. If it takes 400 s to climb the hill, then determine

1. kinetic energy change of the hiker.
2. the potential energy change of the hiker.
3. the work done upon the hiker.
4. the power delivered by the hiker.

1. Delta KE = 0 J
2. Delta PE = +20000 J
3. W = +20000 J
4. P = 50 Watts

a. The speed of the hiker is constant so there is no change in kinetic energy - 0 J.

b. The potential energy change can be found by subtracting the initial PE (0 J) from the final PE (m*g*hf). The final potential energy is 20000 J [from (50 kg)*(10 m/s/s)*(40 m)] and the initial potential energy is 0 J. So Delta PE = +20000 J.

c. The work done upon the hiker can be found using the work-energy theorem. The eq'n reduces to

Wext = PEf

(PEi = 0 J since the hiker starts on the ground; and KEi = KEf since the speed is constant; these two terms can be dropped from the equation since they are equal). The final potential energy is 20000 J [from (50 kg)*(10 m/s/s)*(40 m)]. So W = +20000 J.

d. The power of the hiker can be found by dividing the work by the time.

P=W/t=(20000 J)/(400 s) = 50 Watts

Kinetic Energy (4 seconds) | Potential Energy (12 seconds) | Work (10 seconds) | Power (13 seconds)

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36. Neel, whose mass is 75-kg, ascends the 1.6-meter high stairs in 1.2 s. Determine Neel's power rating. PSYW

Neel'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 75-kg mass up the stairs is determined using the equation

W = F*d*cos(Theta)

where F = m*g = 750 N (approx.), d=1.6 m and Theta = 0 degrees (the angle between the upward force and the upward displacement). Solving for W yields 1200 Joules. Now divide the work by the time to determine the power:

P = W/t = (1200 J)/(1.2 s) = 1000 Watts

Work (10 seconds) | Power (13 seconds)

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37. A 500-kg roller coaster car starts at a height of 32.0 m. Assuming negligible energy losses to friction and air resistance, determine the PE, KE, and speed of the car at the various locations (A, B, C, D, and E) along the track.
 Location Height (m) PE (J) KE (J) velocity (m/s) Start 32.0 160 000 J 0 J 0 A 28.0 140 000 J 20 000 J 8.9 m/s B 11.0 55 000 J 105 000 J 20.5 m/s C 20.0 100 000 J 60 000 J 15.5 m/s D 5.0 25 000 J 135 000 J 23.2 m/s E 15.0 75 000 J 85 000 J 18.4 m/s F 0 0 J 160 000 J 25.3 m/s

The potential energy for every row can be found using the equation PE = m*g*h where m=500 kg and g = 10 m/s/s (approx.). In the first row, the total mechanical energy (KE + PE) equals 160 000 J. Since no work is done by external forces, the total mechanical energy must be 160 000 J in all the other rows. So the KE can be computed by subtracting the PE from 160 000 J. The velocity can be found using the equation:

KE = 0.5*m*v2

where m=500 kg.

Kinetic Energy (4 seconds) | Potential Energy (12 seconds) | Mechanical Energy Conservation (13 seconds)

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38. Use the information in the above table to explain what is meant when it is said that the "total mechanical energy is conserved."

Answer: The total mechanical energy is the same in each row of the table. If the kinetic energy and the potential energy is added together, then the sum will be the same throughout the ride.

Mechanical Energy Conservation (13 seconds)

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39. Use the work-energy theorem to determine the force required to stop a 1000-kg car moving at a speed of 20.0 m/s if there is a distance of 45.0 m in which to stop it. PSYW

The work energy theorem can be written as

KEi + PEi + Wext = 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 + Wext = 0

The expressions for KE (0.5*m*v2) and W (F*d*cos[Theta]) can be sustituted into the equation:

0.5*m*vi2 + F*d*cos[Theta] = 0

where m=1000 kg, vi=20 m/s, d=45 m, and Theta = 180 degrees. Substituting and solving for F yields 4.44*103 N.

Analysis of Situations Involving External Forces (21 seconds) | Application and Practice Questions (10 seconds)

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40. A 60-kg skier accelerates down an icey hill from an original height of 500 meters. Use the work-energy theorem to determine the speed at the bottom of the hill if

1. no energy is lost or gained due to friction, air resistance and other external forces. PSYW
2. 140000 J of energy are lost due to external forces. PSYW

1. v = 100 m/s
2. v = 73.0 m/s

a. Use the work energy theorem:

KEi + PEi + Wext = KEf + PEf

The PEf can be dropped from the equation since the skiier finishes on the ground at zero height. The KEi can also be dropped since the skiier starts from rest. The Wext term is dropped since it is said that no work is done by external forces. The equation simplifies to

PEi =KEf

The expressions for KE (0.5*m*v2) and PE (m*g*h) can be sustituted into the equation:

m*g*h = 0.5*m*vf2

where m=60 kg, h=500 m, g=10 m/s/s (approx.). Substituting and solving for vf yields 100 m/s.

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

m*g*h - 140000J = 0.5*m*vf2

Now substituting and solving for vf yields 73.0 m/s.

Analysis of Situations Involving External Forces (21 seconds) | Mechanical Energy Conservation (13 seconds) | Application and Practice Questions (10 seconds)

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