$$\require{cancel}$$

# 08: Potential Energy and Conservation of Energy

8.1. (4.63 J) − (−2.38 J) = 7.00 J

8.2. 35.3 kJ, 143 kJ, 0

8.3. 22.8 cm. Using 0.02 m for the initial displacement of the spring (see above), we calculate the final displacement of the spring to be 0.028 m; therefore the length of the spring is the unstretched length plus the displacement, or 22.8 cm.

8.4. It increases because you had to exert a downward force, doing positive work, to pull the mass down, and that’s equal to the change in the total potential energy.

8.5. 2.83 N

8.6. F = 4.8 N, directed toward the origin

8.7. 0.033 m

8.8. b. At any given height, the gravitational potential energy is the same going up or down, but the kinetic energy is less going down than going up, since air resistance is dissipative and does negative work. Therefore, at any height, the speed going down is less than the speed going up, so it must take a longer time to go down than to go up.

8.9. Constant U(x) = −1 J 8.10. a. yes, motion confined to −1.055 m ≤ x ≤ 1.055 m ; b. same equilibrium points and types as in example

8.11. x(t) = ± $$\sqrt{\left(\dfrac{2E}{k}\right)} \sin \Big[ \left(\sqrt{\dfrac{k}{m}}\right) t \Big]$$ and v0 = ± $$\sqrt{\left(\dfrac{2E}{m}\right)}$$

# Conceptual Questions

1. The potential energy of a system can be negative because its value is relative to a defined point.

3. If the reference point of the ground is zero gravitational potential energy, the javelin first increases its gravitational potential energy, followed by a decrease in its gravitational potential energy as it is thrown until it hits the ground. The overall change in gravitational potential energy of the javelin is zero unless the center of mass of the javelin is lower than from where it is initially thrown, and therefore would have slightly less gravitational potential energy.

5. the vertical height from the ground to the object

7. A force that takes energy away from the system that can’t be recovered if we were to reverse the action.

9. The change in kinetic energy is the net work. Since conservative forces are path independent, when you are back to the same point the kinetic and potential energies are exactly the same as the beginning. During the trip the total energy is conserved, but both the potential and kinetic energy change.

11. The car experiences a change in gravitational potential energy as it goes down the hills because the vertical distance is decreasing. Some of this change of gravitational potential energy will be taken away by work done by friction. The rest of the energy results in a kinetic energy increase, making the car go faster. Lastly, the car brakes and will lose its kinetic energy to the work done by braking to a stop.

13. It states that total energy of the system E is conserved as long as there are no non-conservative forces acting on the object.

15. He puts energy into the system through his legs compressing and expanding.

17. Four times the original height would double the impact speed.

# Problems

19. 40,000

21. a. −200 J

b. −200 J

c. −100 J

d. −300 J

23. a. 0.068 J

b. −0.068 J

c. 0.068 J

d. 0.068 J

e. −0.068 J

f. 46 cm

25. a. −120 J

b. 120 J

27. a. $$\left(\dfrac{−2a}{b}\right)^{1/6}$$

b. 0

c. ∼ x6

29. 14 m/s

31. 14 J

33. proof

35. 9.7 m/s

37. 39 m/s

39. 1900 J

41. 151 J

43. 3.5 cm

45. 10x with x-axis pointed away from the wall and origin at the wall

47. 4.6 m/s

49. a. 5.6 m/s

b. 5.2 m/s

c. 6.4 m/s

d. No

e. Yes

51. a. where k = 0.02, A = 1, $$\alpha$$ = 1

b. F = kx − $$\alpha xAe^{− \alpha x^{2}}$$

c. The potential energy at x = 0 must be less than the kinetic plus potential energy at x = a or A ≤ $$\frac{1}{2}$$mv2 + $$\frac{1}{2}$$ka2 + $$Ae^{− \alpha a^{2}}$$. Solving this for A matches results in the problem.

53. 8700 N/m

55. a. 70.6 m/s

b. 69.9 m/s

57. a. 180 N/m

b. 11 m

59. a. 9.8 x 103 J

b. 1.4 x 103 J

c. 14 m/s

61. a. 47.6 m

b. 1.88 x 105 J

c. 373 N

63. 33.9 cm

65. a. 0.0269 J

b. U = 0

c. 1.11 m/s

d. 4.96 cm

67. 42 cm

69. 0.44 J

71. 3.6 m/s

73. $$\frac{bD^{4}}{4}$$

75. proof

77. a. $$\sqrt{\dfrac{2m^{2} gh}{k(m + M)}}$$

b. $$\frac{mMgh}{m + M}$$

79. a. 2.24 m/s

b. 1.94 m/s

c. 1.94 m/s

81. 18 m/s

83. vA = 24 m/s

vB = 14 m/s

vC = 31 m/s

85. a. Loss of energy is 240 N • m

b. F = 8 N

87. 89.7 m/s

89. 32 J

# Contributors

• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).