8. b: 88N, 3.83m/s2.
14. Forces: Mg vertically down, N (normal force) perpendicular to the hill and toward it = Mg cos 20o, friction = f, parallel to the hill, up.
Note that if there were no friction there would be a force Mg sin 20o down along the hill and a corresonding acceleration g sin 20o = 3.35m/s2. 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 + Mv2 cos θ = (M) 30. Cancel M. v2 cos θ = 21.8
v2 sin θ = 10 sin 35 = 5.74
v2 = 22.5m/s, Mv2 = 45.1 x 103Kgm/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: v2/r b: If the road is flat and horizontal, the force is the force of friction.
24. a: h = (√ 2 - 1)RE. 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.
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, Mv2/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: Fy = F1 -F3 + F2 sin θ, Fx = F2 cos θ.
14. d: 510 Kg
16. b: 2N, c: O.2 m/s2.
20. yes, no, yes.
22. a: The force of friction. b: 2.25
30. v and a are the same.
32. a: 17.06N, c: θ = 6.56o, d: it would remain the same.