171 Problem Set 11

Kinetic and Potential Energy
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This problem set deals with kinetic and potential energy.
1.

Match each description of energy given in a to f below with one of the types of energy:

  1. gravitational potential energy
  2. elastic potential energy
  3. kinetic energy
  1.  1  Water stored behind a dam.
  2.  3  An avalanche in progress.
  3.  2  An auto brake return spring.
  4.  1  A book standing on a shelf.
  5.  1  A pendulum at the top of its swing.
  6.  3  A pendulum at the bottom of its swing.
2.

Given the following relationships, fill in the blanks to define the symbols.

a. Ep = w · h where w = weight in lb, oz, N
h = height in ft, in, m
     
b. Ep = (½)kd2 where k = spring constant in lb/in, lb/ft, N/cm, N/m
d = distance from neutral position in in, ft, cm, m
 
3. What is the average potential energy of a cloud centered at 5000 ft if the condensed moisture in the cloud has a total weight of 2100 lb?

  
Knowns Unknowns Equations/Remarks
h = 5000 ft
w = 2100 lb		 Ep = ? Ep = wh 

Ep = wh = (2100 lb)(5000 ft)

Ep = 1.05 ×107 ft·lb

Substitute in the known values and calculate the answer.
4. A spring is compressed 2 in by a force of 10 lb. What is the spring constant, k, of this spring?

  
Knowns Unknowns Equations/Remarks
d = 2 in
F = 10 lb		 k = ? F = kd 

F = kd 

k = F/d = (10 lb)/(2 in)

k = 5 lb/in

Rearrange the equation to solve for the spring constant, substitute in the known values, and calculate the answer.
5. The work done to lift or move an object some height above a reference point is stored as gravitational potential energy.  
6. How much stored potential energy is released when 1000 lb of water flow out of a factory fire sprinkler system storage tank located 150 ft above the factory floor?

  
Knowns Unknowns Equations/Remarks
w = 1000 lb
h = 150 ft		 Ep = ? Ep = wh 

Ep = wh = (1000 lb)(150 ft)

Ep = 150,000 ft·lb

Substitute in the known values and calculate the answer.
7. How much potential energy is gained when 50 m3 of water are pumped into a storage tank 50 ft above the ground? (r = 1000 kg/m3)

 
Knowns Unknowns Equations/Remarks
V = 50 m3
h = 50 ft
r = 1000 kg/m3
g = 9.8 N/kg
1 m = 3.28 ft		 Ep = ? Ep = mgh 
r = m/V 

r = m/V 
m = rV

Ep = mgh = rVgh 
Ep = (1000 kg/m3)(50 m3)(9.8 N/kg)(50 ft)(1 m/3.28 ft)
Ep = 7.47 ×106 J

Rearrange the density equation to find the mass and substitute in the potential energy equation.

Convert the feet into meters. Note that using 9.8 N/kg instead of m/s2 for gravity makes for easier cancellation of units. Substitute in the known values and calculate the answer.
8.

A newspaper printing machine uses a set of rollers to feed paper through the machine. The rollers are held firmly against the paper by a spring. The spring exerts a force of 100 lb when it is compressed 4 in beyond its unstretched length.

  1. Find the spring constant.
  2. Find the amount of work done to compress the spring 4 inches.
  
Knowns Unknowns Equations/Remarks
F = 100 lb
d = 4 in
1 ft = 12 in		 a. k = ?
b. W = ?		 F = kd 
W = ½kd2 

a. F = kd 
    k = F/d = (100 lb)/(4 in)
    k = 25 lb/in

b. W = ½kd2 = ½(25 lb/in)(4 in)2(1 ft/12 in)
    W = 16.7 ft·lb

Rearrange the equation to solve for the spring constant, substitute in the known values, and calculate the answer.

Substitute in the known values, convert inches to feet, and calculate the answer.
9.

Given that 500 in·lb of work is done in stretching a spring with a spring constant of 40 lb/in, find

  1. the distance the spring is stretched, and
  2. the force applied to stretch the spring.
  
Knowns Unknowns Equations/Remarks
W = 500 in·lb
k = 40 lb/in		 a. d = ?
b. F = ?		 W = ½kd2 
F = kd 

a. W = ½kd2 
    d = (2W/k)½
    d = [(2)(500 in·lb)/(40 lb/in)]½
    d = 5 in

b. F = kd = (40 lb/in)(5 in)
    F = 200 lb

Rearrange the work equation to solve for the distance the spring is stretched.

Use this distance in the spring equation to find the force applied to the spring.
10.

A hydraulic accumulator spring with a spring constant of 200 lb/in is compressed 6 inches.

  1. Find the potential energy stored in the spring when it is compressed 6 in.
  2. Find the force required to compress the spring 6 inches.
  
Knowns Unknowns Equations/Remarks
k = 200 lb/in
d = 6 in		 a. Ep = ?
b. F = ?		 Ep = ½kd2 
F = kd 

a. Ep = ½kd2 
    Ep = (1/2)(200 lb/in)(6 in)2
    Ep = 3600 in·lb

b. F = kd = (200 lb/in)(6 in)
    F =  1200 lb

Substitute in the known values and calculate the answers.

 
11.

After a hail storm, the weather bureau reported that the hail had a speed of 550 ft/sec on impact and fell from a cloud 4800 ft above the ground.

  1. If the average hailstone weighed 0.175 lb, how much energy was involved in the damage caused by one hailstone?
  2. What type of energy was involved at the moment the hailstone struck the ground?
 
Knowns Unknowns Equations/Remarks
v = 550 ft/sec
h = 4800 ft
w = 0.175 lb		 a. E = ?
b. What type of energy?		 Ep = wh 

a. Ep = wh = (0.175 lb)(4800 ft)
    Ep = 840 ft·lb

b. The energy involved was kinetic energy.

The potential energy that the hailstone had at 4800 ft is converted into kinetic energy. We can use the kinetic energy equation and get an answer of 827 ft·lbs, which differs from the answer in the book at a percent error of 1.9 %..

12.

A motorcycle front wheel spring shock has a spring constant k = 300 lb/in. The spring is compressed 2 in beyond its normal rest position while going over a rock. How much potential energy is stored in the spring?

  
Knowns Unknowns Equations/Remarks
k = 300 lb/in
d = 2 in		 Ep = ? Ep = ½kd2 

Ep = ½kd2 = (1/2)(300 lb/in)(2 in)2

Ep = 600 in·lb

Substitute in the known values and calculate the answers.
13.

A 20 lb object is resting on a bench top 3 ft above the floor. It is moved to a shelf that is 5 ft above the floor.

  1. How much work is done in lifting the object from the bench to the shelf?
  2. What is the potential energy of the object sitting on the shelf with respect to the bench top?
  3. What is its potential energy with respect to the floor?
  4. What would it mean for an object to have a negative potential energy?
  
Knowns Unknowns Equations/Remarks
w = 20 lb
h1 = 3 ft
h2 = 5 ft		 a. W = ?
b. Ep = ? shelf to benchtop
c. Ep = ? shelf to floor
d. negative Ep means what?		 W = wh 
Ep = wh 

a. W = w(h2 – h1)
    W = (20 lb)(5 ft – 3 ft)
    W = 40 ft·lb

b. Ep = W = 40 ft·lb

c. Ep = wh2 = (20 lb)(5 ft)
    Ep = 100 ft·lb  

d. Negative potential energy means that the level the object is at is below the reference level (which is usually the ground).

The work done lifting the object is its weight times the vertical displacement (h2 – h1).

The gravitational potential energy of the object on the shelf with respect to the benchtop is the amount of work that was required to move the object to the shelf.

The ground is the reference level and its height is zero.
14. A 15 kg container is sitting on a shelf 3.770 m high. Find the gravitational potential energy of the container with respect to the floor.

  
Knowns Unknowns Equations/Remarks
m = 15 kg
h = 3.770 m
g = 9.8 N/kg		 Ep = ? Ep = mgh 

Ep = mgh 

Ep = (15 kg)(9.8 N/kg)(3.770 m)

Ep = 554 J

Substitute in the known values and calculate the answers.
15.

A spring is compressed by 4 inches, requiring a force of 12.0 lb.

  1. What is the average applied force on the spring during the compression?
  2. How much work is done on the spring by the average applied force?
  3. Find the spring constant in lb/ft.
  4. Calculate the elastic potential energy of the compressed spring.
  5. What is the relationship between the answers to (b) and (d)? Explain.
  
Knowns Unknowns Equations/Remarks
d = 4 in
F = 12.0 lb
1 ft = 12 in		 a. Fav = ?
b. W = ?
c. k = ?
d. Ep = ?
e. What is relationship between (b) and (d)?		 Fav = F/2 
W = Fav·d 
k = F/d 
Ep = ½kd2 

a. Fav = F/2 = (12.0 lb)/2
    Fav = 6 lb  

b. W = Fav·d = (6 lb)(4 in)(1 ft/12 in)
    W = 2 ft·lb 

c. k = F/d = (12.0 lb)/[(4 in)(1 ft/12 in)]
    k = 36 lb/ft

d. Ep = ½kd2 = (1/2)(36 lb/ft)[(4 in)(1 ft/12 in)]2
    Ep = 2 ft·lb

e. The answers to (b) and (d) are equal because the potential energy of the spring is the work done to compress the spring.

 
16.

A 10 cm spring has a Hooke's Law constant of 200 N/m.

  1. How much potential energy will it have when it is compressed from 10 cm to 9 cm?
  2. How much potential energy will it have when it is compressed an additional one centimeter?
  3. Compare the potential energy of the first centimeter of compression with the gain in potential energy caused by the second centimeter of compression. Why are these values not the same?
  
Knowns Unknowns Equations/Remarks
k = 200 N/m
d1 = 1 cm
d2 = 2 cm
1 m = 100 cm		 a. Ep1 = ?
b. Ep2 = ?
c. Compare the two Ep's		 Ep = ½kd2 

a. Ep1 = ½kd12 = (1/2)(200 N/m)[(1 cm)(1 m/100 cm)]2
    Ep1 = 0.01 J

b. Ep2 = ½kd22 = (1/2)(200 N/m)[(2 cm)(1 m/100 cm)]2
    Ep2 =  0.04 J

c. The change in potential energy for the first cm is 0.01 J, while the change for the second cm is 0.03 J. This difference is due to the increasing amount on force needed to compress the spring an additional distance.

Find the potential energy of the spring in both cases.

In the second case, the change in displacement is equal to the first case (1 cm), but the change in potential energy is not the same.

17.

A spring with a Hooke's Law constant of 6 lb/in is stretched.

  1. Determine the elongation of the spring for applied forces of 24 lb and 36 lb.
  2. Plot a graph of force versus elongation using these two points. On your graph let one square represent 12 lb in the vertical direction and one square represent one inch in the horizontal direction.
  3. Draw a straight line through the two points. Should the line go through the origin?
  4. How much energy is represented by the area of one square?
  5. Find the area under your line, from the origin to the force value of 36 lb. How much energy is indicated?
  6. Use the equation for elastic potential energy of a spring to calculate the elastic potential energy. Are the answers to (e) and (f) equal?
  
   
Knowns Unknowns Equations/Remarks
k = 6 lb/in
F1 = 24 lb
F2 = 36 lb
1 ft = 12 in		 a. d1 and d2 = ?
b. Plot a graph of F vs d
c. Draw line
d. Energy of 1 square = ?
e. A under line = ?
f. Ep = ? Are (e) and (f) equal?		 k = F/d 
W = Fd 
A = bh/2 (area of a triangle)
Ep = ½kd2 

a. k = F/d 
    d1 = F1/k = (24 lb)/(6 lb/in) = 4 in
    d2 = F2/k = (36 lb)/(6 lb/in) = 6 in

c. Yes, the line should go through the origin. With zero force applied, the displacement is zero.

d. W = Fd = (12 lb)(1 in)(1 ft/12 in) = 1 ft·lb

e. A = bh/2 = (1/2)(6 sq)(3 sq)(1 ft·lb/1 sq2) = 9 ft·lbs

f. Ep = ½kd2 = (1/2)(6 lb/in)(6 in)2(1 ft/12 in)
    Ep =  9 ft·lbs. Yes, the answers to (e) and (f) are equal.

a. Rearrange the spring constant equation to solve for the displacement.

d. Force times distance is work (energy).

e. The area under the line is 9 squares. Each square is 1 ft·lb of energy, so the energy is 9 ft·lbs.

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