1. $(U{}_{\text{G}}){}_{\text{C}}>(U{}_{\text{G}}){}_{\text{B}}=(U{}_{\text{G}}){}_{\text{D}}>(U{}_{\text{G}}){}_{\text{A}}$. Gravitational potential energy depends only on height, not on speed.

2. b. Potential energy depends only on the vertical displacement. At the elevation of the dashed line, both have gained the same gravitational potential energy, so both have lost the same kinetic energy.

3. $v{}_{\text{A}}=v{}_{\text{B}}=v{}_{\text{C}}=v{}_{\text{D}}$. Her increase in kinetic energy depends only on the vertical distance through which she falls, not on the shape of the slide.

4. c. Constant speed means no change of kinetic energy. But for motion on a slope, constant speed requires friction. All the gravitational potential energy is being transformed into thermal energy.

5. c. $U_{\text{Sp}}$ depends on $d^2$. Doubling the compression increases $U_{\text{Sp}}$ by a factor of 4. All potential energy is transformed into kinetic energy, so $K$ increases by a factor of 4. But $K$ depends on $v^2$, so $v$ increases by only a factor of 2.

6. $x=6\mathbf{\text{m}}\mathbf{\text{.}}$ From the graph, the system’s potential energy when the particle is at $x=1\text{m}$ is $U=3\text{J}$. Its total energy is thus $E=K+U=4\text{J}$. A TE line at 4 J crosses the PE curve at $x=6\text{m}$.

7. d. The system is losing potential energy as the weight falls. It’s gaining speed, so some $U$ is transformed into $K$. Energy is also being transferred out of the system to the environment via negative work done by the rope tension. The system is not isolated, so neither $E_{\text{mech}}\text{nor}{E}_{\text{sys}}$ is conserved.

1.

1. 5 J

2. 7 J

3. $\text{Δ}U=2.9\times {10}^6\text{J}$

5.

1. $13\text{ m}\text{/}\text{s}$

2. $14\text{ m}\text{/}\text{s}$

7. 1.0 m

9. $4.2\text{ m}\text{/}\text{s}$

11. $1.4\text{ m}\text{/}\text{s}$

13. $81\text{ m}\text{/}\text{s}$

15. 0.13 J

17. 18 J

19. $60\text{ m}\text{/}\text{s}$

21. $0.71\text{ m}\text{/}\text{s}$

23. 9.7 J

25.

1. Right

2. 17 $\text{m}\text{/}\text{s at }x=4\text{ m}$

3. $x=1\text{m}$, $x=6\text{m}$

27. $63\text{ m}\text{/}\text{s}$

29. $6.3\text{ m}\text{/}\text{s}$

31. 100 N at $x=5\text{cm}$, 0 N at $x=15\text{cm}$, $-50\text{N at x}=25\text{cm}$

33. 2.5 N, 0.40 N, 0.16 N

35.

1. Yes

2. 20 J

37.

39. $-1\text{J}$ of work is done to the environment.

41. $5.4\text{ m}\text{/}\text{s}$

43.

1. $2.2\times {10}^4\text{N}\text{/}\text{m}$

2. $19\text{ m}\text{/}\text{s}$

45. 43 m

47. 65 g

49.

1. $v_{\text{f}}=\sqrt{{\displaystyle \frac{2gh(m_{\text{B}}-m_{\text{A}})}{(m_{\text{B}}+m_{\text{A}})}}}$

2. 0.50 m

51.

1. $v_{\text{f}}=\sqrt{{\displaystyle \frac{2gh}{M+m}}(m-{\mu }_{\text{k}}M)}$

2. $v_{\text{f}}=\sqrt{{\displaystyle \frac{2gmh}{M+m}}}$

53.

1. 0.51 m

2. 0.38 m

55.

1. $3\times {10}^3\text{m}\text{/}\text{s}$

2. $2\times {10}^1\text{THz}$

57. $3.3\text{ m}\text{/}\text{s}$

59.

1. 0.5 m, 1.5 m, 2.5 m

2. Stable, unstable, stable

61.

1. $\frac{2A}{B}$

2. Both are stable.

63.

1. $-\pi B/L$

2. $-AL$

3. $-2AL+\pi B/L$

65.

1. $a^{2}b$

2. $a^{2}b$

3. Yes

67. $15\text{ m}\text{/}\text{s}$

69. $2.6\text{ m}\text{/}\text{s}$

71. $20\text{ N}\text{/}\text{m}$

73.

1. 1.5 m

2. 20 cm

75. $1.37\times {10}^8\text{m}\text{/}\text{s}$