### Answers and comments for the review activities and problems in Chapter 4.

#### GR

3. 54.4Kg, 533N
5. 7.17m/s

8. b: 88N, 3.83m/s^{2}.

14. Forces: Mg vertically down, N (normal force) perpendicular to the hill and toward it = Mg cos 20^{o}, friction = f, parallel to the hill, up.

Note that if there were no friction there would be a force Mg sin 20^{o} down along the hill and a corresonding acceleration g sin 20^{o} = 3.35m/s^{2}. Since this is larger than the actual (given) acceleration, there is friction. To find it and the other forces we need to know the mass.

16. a: The only force is Mg vertically down.

b: ('is' should be 'its'.) Horizontal. c: g vertically down.

18. a: 15m/s

b: (M)10 cos 35 + Mv_{2} cos θ = (M) 30. Cancel M. v_{2} cos θ = 21.8

v_{2} sin θ = 10 sin 35 = 5.74

v_{2} = 22.5m/s, Mv_{2} = 45.1 x 10^{3}Kgm/s.

20. If you are pushing to the right the forces are 133.4N to the right and 66.7N to the left.

23. a: v^{2}/r b: If the road is flat and horizontal, the force is the force of friction.

24. a: h = (√ 2 - 1)R_{E}. b: 1.81h

27. Each of these atoms has just a single electron outside a closed shell of electrons. The spin angular momentum of a closed shell of electrons is zero.

#### P&RSB

1. Without a horizontal force on him or her, the passenger would move in a straight line with constant velocity. In order for the passenger to move in a circle there must be a force, the centripetal force, toward the center of the motion.

2. If the car is in contact with the track, there are two forces at the top: the car's weight, Mg, and the force of the track on the car, N, both vertically down. Their sum is the centripetal force, Mv^{2}/R. The minimum speed for the car to remain in contact with the track is that for which N = 0, so that the centripetal force is Mg and the speed is √ (gR).

6. a: 9m, b: The force is now -2N, causing the block to decelerate until it stops. Extra question: how much further will the block move before it stops, and how long will that take? (Ans.: 36m, 12s.)

10. a: F_{y} = F_{1} -F_{3} + F_{2} sin θ, F_{x} = F_{2} cos θ.

14. d: 510 Kg

16. b: 2N, c: O.2 m/s^{2}.

20. yes, no, yes.

22. a: The force of friction. b: 2.25

24. 0.061%

30. v and a are the same.

32. a: 17.06N, c: θ = 6.56^{o}, d: it would remain the same.

#### MCQ

2. b

7. c