Chapter 8 Potential Energy and Conservation of …physicsweb.phy.uic.edu/105/JC/08_LectureOutline.pdf · Chapter 8: Potential Energy and Conservation of Energy James S. Walker,

Copyright © 2010 Pearson Education, Inc.

Chapter 8Potential Energy and

Conservation of Energy


8-1 Conservative and Nonconservative Forces

Conservative force: the work it does is stored in the form of energy that can be released at a later timeExample of a conservative force: gravityExample of a nonconservative force: frictionAlso: the work done by a conservative force moving an object around a closed path is zero;Example: paint can in Problem 7.8.This is not true for a nonconservative force



Work done by gravity on a closed path is zero:



Work done by friction on a closed path is not zero:



The work done by a conservative force is zero on any closed path:


8-2 The Work Done by Conservative Forces

If we pick up a ball and put it on the shelf, we have done work on the ball. We can get that energy back if the ball falls back off the shelf; in the meantime, we say the energy is stored as potential energy.

(8-1)



Gravitational potential energy:



Springs: (8-4)


9


10

Chapter 8: Potential Energy and Conservation of Energy James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8 – 5

12 Picture the Problem: The Hawkmoth wing behaves as an ideal spring.

Strategy: Use Hooke’s Law (equation 6-4) to find the force constant of the wing, then use equation 8-5 to find the energy stored. Combine both equations to find the force required to store twice the energy.

Solution: 1. (a) Solve equation 6-4 for k: 0.0030 N0.625 N/m 0.63 N/m

0.0048 mFkx

� � �

2. (b) Use equation 8-5 to find U: � �� 22 61 12 2 0.625 N/m 0.0048 m 7.2 10 J 7.2 JU kx P� u

3. (c) Solve equation 8-5 for x, taking the negative root because the wing is depressed in the negative direction as in step 1:

22 2

U Ux xk k

� �

4. Use Hooke’s Law to find the force: � �� 6

2 2

2 2 7.2 10 J 0.625 N/m 0.0042 N 4.2 mN

F kx k U k Uk

�

� � �

u u

Insight: We bent the rules for significant figures for k a bit in steps 2 and 4 in order to avoid rounding error. Notice that in step 4 the force required to double the stored energy is 2 bigger than the original force.

13. Picture the Problem: A mass is suspended from a vertical spring. As the spring is stretched it stores potential energy.

Strategy: Doubling the mass doubles the force exerted on the spring, and therefore doubles the stretch distance due to Hooke’s Law .F kx � Use a ratio to find the increase in spring potential energy 21

2U kx when the stretch distance is

doubled.

Solution: 1. (a) If the mass attached to the spring is doubled the stretch distance will double. Doubling the extension of the spring will increase the potential energy by a factor of 4.

2. (b) Doubling the mass doubles the force and therefore doubles the stretch distance:

1

2 12 2

x mg kx mg k x

3. Calculate the ratio 2 1U U : � �

� �

221122 2

2 211 1 12

2 1

24

4 4 0.962 J 3.85 J

xkxUU kx x

U U

Insight: Note that the change in gravitational potential energy also quadruples as the new mass is hung on the spring. It doubles because there is twice as much mass and it doubles again because the spring stretches twice as far. If you calculate the stretch distances you’ll find the spring stretched 1 1 12 5.6 cmx U m g when the 3.5-kg mass was suspended and it stretched 11.2 cm when the 7.0-kg mass was suspended from it.

14. Picture the Problem: The spring in the soap dispenser is compressed by the applied force.

Strategy: Use equation 8-5 to find the spring constant using the given energy and compression distance data. Solve the same equation for x in order to answer part (b).

Solution: 1. (a) Solve equation 8-5 for k:

� �� 2 2

2 0.0025 J2200 N/m 0.20 kN/m

0.0050 m

Ukx

2. (b) Solve equation 8-5 for x:

� �2 0.0084 J20.92 cm

200 N/mUxk

Insight: To compress the spring of this dispenser 0.50 cm requires 1.0 N or about ¼ lb of force.


8-3 Conservation of Mechanical EnergyDefinition of mechanical energy:

(8-6)

Using this definition and considering only conservative forces, we find:

Or equivalently:


8-3 Conservation of Mechanical EnergyEnergy conservation can make kinematics problems much easier to solve:


13


14

Chapter 7: Work and Kinetic Energy James S. Walker, Physics, 4th Edition


7 – 6

21. Picture the Problem: The bullet moves at high speed in a straight line.

Strategy: Calculate the kinetic energy using equation 7-6, then use ratios to find the new kinetic energies in (b) and (c).

Solution: 1. (a) Apply equation 7-6 directly: � �� 221 12 2 0.00950 kg 1300 m/s 8030 J 8.03 kJK mv

2. (b) Use a ratio to predict the new kinetic energy:

� �

� �

22 11old2newnew 2

2 21old old old2

1 1new old4 4

1 so that4

8.03 kJ 2.01 kJ

vmvKK mv v

K K

3. (c) Use a ratio to predict the new kinetic energy:

� �

� �

221oldnewnew 2

2 21old old old2

new old

24 so that

4 4 8030 J 32.1 kJ

vmvKK mv v

K K

Insight: The kinetic energy is proportional to the velocity squared, a useful thing to remember when calculating ratios. 22. Picture the Problem: The work W0 accelerates a car from 0 to 50 km/h. Strategy: Use the work-energy theorem to answer the conceptual question.

Solution: 1. (a) The work done on the car is equal to its change in kinetic energy, which in turn is proportional to the square of its velocity. Therefore, if the kinetic energy of the car is W0 when its speed is 50 km/h, its kinetic energy will be 9W0 when its speed is 150 km/h. We conclude that the work required to accelerate the car from 50 km/h to 150 km/h is 9W0 í W0 = 8W0.

2. (b) The best explanation is II. The final speed is three times the speed that was produced by the work W0. Statement I is false because the kinetic energy depends on the speed squared, and the work is proportional to the changein kinetic energy. Statement III is true but irrelevant.

Insight: By using a similar analysis you can show that a car that is traveling 35 mph has twice the kinetic energy as a car that is traveling 25 mph. In other words, it takes just as much work to increase the car’s speed from 25 to 35 mph as it did to accelerate from 0 to 25 mph.

23. Picture the Problem: Four joggers have a variety of masses and speeds.

Strategy: Use the definition of kinetic energy to determine the relative magnitudes of the kinetic energies.

Solution: 1. Calculate the kinetic energies of each jogger.

2. � �� 2 21 1A 2 2K m v mv 3. � �� 2 291 1 1

B 2 2 2 23K m v mv

4. � �� 2 231 1 1C 2 2 4 23K m v mv 5. � �� 2 21 1 1

D 2 2 24K m v mv

6. By comparing the magnitudes of the kinetic energies we arrive at the ranking C < A = D < B.

Insight: Even with three times the mass of jogger A, jogger C has only three-fourths the kinetic energy because the kinetic energy is proportional to the square of the speed.

24. Picture the Problem: The pine cone falls straight down under the influence of gravity.

Strategy: The work done by gravity equals the change in kinetic energy according to equation 7-7. The work done by gravity is always W = mgh as indicated in Example 7-2 and Conceptual Checkpoint 7-1.

Solution: 1. (a) The work done by gravity on the pine cone equals the increase in its kinetic energy. Set the energies equal and solve for v: � ��

212

22 2 9.81 m/s 16 m 18 m/s

W K mgh mv

v gh

'

2. (b) Air resistance did negative work because the speed and therefore the kinetic energy of the pine cone when it landed was reduced. Air resistance removed energy from the pine cone.

Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion.


8-4 Work Done by Nonconservative Forces

In this example, the nonconservative force is water resistance:


8-4 Work Done by Nonconservative ForcesIn the presence of nonconservative forces, the total mechanical energy is not conserved:

Solving,

(8-9)


17



8 – 3

2. Sum the work done by the spring for each segment of path 2:

� � � �� ^ `

� � � �

2 2 2 212 1 4 4 32

2 22 212

2

550 N/m 0 0.020 m 0.020 m 0.020 m

0.1 J 0 J 0.11 J

W k x x x x

W

ª º � � �¬ ¼

ª º ª º � � � � �¬ ¼ ¬ ¼

� � �

3. (b) The work done by the spring will stay the same if you increase the mass because the results do not depend on the mass of the block.

Insight: The work done by the spring is negative whenever you displace the block away from x = 0, but it is positive when the displacement vector points toward x = 0.

5. Picture the Problem: The two paths of the object are shown at right.

Strategy: The work done by gravity is W mgy when the object is moved downward, W mgy � when it is moved upward, and zero when it is moved horizontally. Sum the work done by gravity for each segment of each path.

Solution: 1. (a) Calculate the work for path 1: � ��

1 12

05.2 kg 9.81 m/s 1.0 m 51 J

W mg y �

2. Calculate the work for path 2: � ��

2 12

05.2 kg 9.81 m/s 1.0 m 51 J

W mg y �

3. (b) If you increase the mass of the object the work done by gravity will increase because it depends linearly on m.

Insight: The work is path-independent because gravity is a conservative force. 6. Picture the Problem: The physical situation is depicted at right.

Strategy: Use � �2 21sp i f2W k x x � for the work done by the spring.

That way the work will always be negative if you start out at i 0x because the spring force will always be in the opposite direction

from the stretch or compression. The work done by kinetic friction is fr kW mgdP � , where d is the distance the box is pushed irregardless

of direction, because the friction force always acts in a direction opposite the motion.

Solution: 1. (a) Sum the work done by the spring for each segment of path 2:

� � � �� ^ `

� � � �

2 2 2 21sp 1 4 4 32

2 22 212

sp

480 N/m 0 0.020 m 0.020 m 0.020 m

0.096 J 0 J 0.096 J

W k x x x x

W

ª º � � �¬ ¼

ª º ª º � � � � �¬ ¼ ¬ ¼

� � �

2. Sum the work done by friction for each segment of path 2:

� ��

fr k 1 2

20.16 2.7 kg 9.81 m/s 0.020 0.040 m 0.25 J

W mg d dP � �

� � �

3. (b) Sum the work done by the spring for the direct path from A to B:

� ��

2 21sp A B2

2212 480 N/m 0 0.020 m 0.096 J

W k x x �

ª º � �¬ ¼

4. Sum the work done by friction for the direct path from A to B:

� �� 2fr k 0.16 2.7 kg 9.81 m/s 0.020 m 0.085 JW mgdP � � �

Insight: The work done by friction is always negative, and increases in magnitude with the distance traveled.

Calculate total work from A to Bk = 480 N/m; m = 2.7 kg; µ = 0.16

1) along path 2

2) Directly from A to B


18



8 – 3

2. Sum the work done by the spring for each segment of path 2:

� � � �� ^ `

� � � �

2 2 2 212 1 4 4 32

2 22 212

2

550 N/m 0 0.020 m 0.020 m 0.020 m

0.1 J 0 J 0.11 J

W k x x x x

W

ª º � � �¬ ¼

ª º ª º � � � � �¬ ¼ ¬ ¼

� � �

3. (b) The work done by the spring will stay the same if you increase the mass because the results do not depend on the mass of the block.

Insight: The work done by the spring is negative whenever you displace the block away from x = 0, but it is positive when the displacement vector points toward x = 0.

5. Picture the Problem: The two paths of the object are shown at right.

Strategy: The work done by gravity is W mgy when the object is moved downward, W mgy � when it is moved upward, and zero when it is moved horizontally. Sum the work done by gravity for each segment of each path.

Solution: 1. (a) Calculate the work for path 1: � ��

1 12

05.2 kg 9.81 m/s 1.0 m 51 J

W mg y �

2. Calculate the work for path 2: � ��

2 12

05.2 kg 9.81 m/s 1.0 m 51 J

W mg y �

3. (b) If you increase the mass of the object the work done by gravity will increase because it depends linearly on m.

Insight: The work is path-independent because gravity is a conservative force. 6. Picture the Problem: The physical situation is depicted at right.

Strategy: Use � �2 21sp i f2W k x x � for the work done by the spring.

That way the work will always be negative if you start out at i 0x because the spring force will always be in the opposite direction

from the stretch or compression. The work done by kinetic friction is fr kW mgdP � , where d is the distance the box is pushed irregardless

of direction, because the friction force always acts in a direction opposite the motion.

Solution: 1. (a) Sum the work done by the spring for each segment of path 2:

� � � �� ^ `

� � � �

2 2 2 21sp 1 4 4 32

2 22 212

sp

480 N/m 0 0.020 m 0.020 m 0.020 m

0.096 J 0 J 0.096 J

W k x x x x

W

ª º � � �¬ ¼

ª º ª º � � � � �¬ ¼ ¬ ¼

� � �

2. Sum the work done by friction for each segment of path 2:

� ��

fr k 1 2

20.16 2.7 kg 9.81 m/s 0.020 0.040 m 0.25 J

W mg d dP � �

� � �

3. (b) Sum the work done by the spring for the direct path from A to B:

� ��

2 21sp A B2

2212 480 N/m 0 0.020 m 0.096 J

W k x x �

ª º � �¬ ¼

4. Sum the work done by friction for the direct path from A to B:

� �� 2fr k 0.16 2.7 kg 9.81 m/s 0.020 m 0.085 JW mgdP � � �

Insight: The work done by friction is always negative, and increases in magnitude with the distance traveled.


8-5 Potential Energy Curves and Equipotentials

The curve of a hill or a roller coaster is itself essentially a plot of the gravitational potential energy:



The potential energy curve for a spring:



Contour maps are also a form of potential energy curve:

Documents

Chapter 8 Potential Energy and Conservation of …physicsweb.phy.uic.edu/105/JC/08_LectureOutline.pdf · Chapter 8: Potential Energy and Conservation of Energy James S. Walker,