1. The crate’s gravitational force and the normal force are incorrectly identified as an action/reaction pair. The normal force should be paired with a downward force of the crate on the ground. Gravity is the pull of the entire earth, so $\vec{F}{}_{\text{G}}$ should be paired with a force pulling up on the entire earth.

2. c. Newton’s third law says that the force of A on B is equal and opposite to the force of B on A. This is always true. The mass of the objects isn’t relevant.

3. b. $F_{\text{B on H}}=F_{\text{H on B}}\text{ and }{F}_{\text{A on B}}=F_{\text{B on A}}$ because these are action/reaction pairs. Box B is slowing down and therefore must have a net force to the left. So from Newton’s second law we also know that $F_{\text{H on B}}>F_{\text{A on B}}$.

4. Equal to. Each block is hanging in equilibrium, with no net force, so the upward tension force is $mg$.

5. Less than. Block B is accelerating downward, so the net force on B must point down. The only forces acting on B are the tension and gravity, so $T_{\text{S on B}}<(F_{\text{G}}){}_{\text{B}}$.

1.

• a. and b.

  

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• b. The system is the racket and the ball.

• c.

  

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3.

• a.

  

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• b. The system is the cable and the girder.

• c.

  

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5.

• a.

  

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• b. The system is block A and block B.

• c.

  

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7.

1. $7.8\times {10}^2\text{N}$

2. $1.6\times {10}^3\text{N}$

9.

1. 6 N

2. 10 N

11. 5.0 kg

13. 42 m

15. 270 N

17.

1. 32 N

2. 19 N

3. 16 N

4. 3.2 N

19. 3.9 kN

21. 67 N, $36\text{°}$

23.

1. 3.9 N

2. $2.2{\text{ m/s}}^2$

25. 0.86 kg

27.

• a.

  

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• b. 4.0 N

29.

1. 1.3 kg

2. 1.4 kg

31. 99 m

33. 1.5 s

35. 1.8 s

37. $2.3{\text{ m/s}}^2$

39.

1. 1.8 kg

2. $1.3{\text{ m/s}}^2$

41.

1. 20 N

2. Down the slope

3. 21 N

43.

1. $8.2\times {10}^3\text{N}$

2. $4.8\times {10}^2\text{N}$

45. $4.0\times {10}^2\text{N}$

47. 920 g

49. $F=(m_1+m_2)g\text{tan}\theta$

53. $1.8\text{m}\text{/}{\text{s}}^2$

55. $2.8\text{m}\text{/}{\text{s}}^2$

57. $2.4\times {10}^{-6}\text{N}$

59. $t_{1/2}={\displaystyle \frac{\text{ln}\left(2\right)}{b({\displaystyle \frac{1}{m}}+{\displaystyle \frac{1}{M}})}}$