Page 58 - SP015 Past Years PSPM Chapter 1 -5 Ver.2020
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PHYSICS PSPM SEM 1 1999 - 2017
FINAL ANSWER FOR CHAPTER 4: FORCES
1. Mass is a property of matter and weight is a force due to gravity.
2.
5. Newton’s First Law states that a body will remain at rest or at constant velocity in a straight line unless
it is forced to change its condition by the forces acting on it.
6. Equilibrium of particle is when the vector sum of force that acts on the particle equals zero.
F 0
Equlibrium of a particle is a condition when the net force acting on the particle is zero.
The equilibrium of forces means that the total force is zero @ F 0 .
7. Static eqilibrium and dynamic equilibrium.
8. Ratio of tension = 2:3
9. F = 13 N at –22.6 from negative x-axis @ 157.4 from x-axis
10. R = 12.68 N
11. F 3 = 1304 N at –15.57 from negative x-axis or 164.43 from x-axis
12. F = 3.33 N at 28.14 from x-axis
13. = 0.59
14. (b) T 1 = 58.9 N (c) m C = 21.2 kg
15. (b) F = 2.3 N
(c) The eraser will slide upward because F 1 > F thus, net upward force exists.
16. N = 490.5 N
17. (a) No force is needed to maintain an object moving with a constant velocity.
v = constant a = 0 F m a 0
(b) Draw a free body diagram with correct label and arrow.
f s = mg sin and N = mg cos
18. Since F x 0 and F y 0 or net force or resultant force exist thus particle is not in equilibrium.
19. (b) F = 6.94 N
20. (b) T P = 162.4 N, T Q = 68.6 N
21. T 1 = 45.31 N, T 2 = 22.65 N
22. (b) Mass of block Q, M Q = 1.02 kg
23. Newton’s second law of motion states that the net applied force is proportional to the rate of change of
momentum and is in the direction of the straight line along which the force acts.
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24. 1 N is the net force on a 1 kg mass that causes it to accelerate 1 m s .
25. An object thrown upward.
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26. (a) max = 21. 8 (b) a = 0.91 m s
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27. T = 42.16 N, a = 4.84 m s
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28. (a) a = –1 m s (b) = 5.9
29. = 0.133
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30. (a) Copy and label all the forces acting on both blocks. (b) a = 1.83 m s (c) T = 23.94 N
31. (b) T = 3.91 N
32. F = 15000 N @ F = –15000 N
4
4
33. (a) T = 1.42 10 N (b) T = 1.18 10 N
34. (b) = 0.72
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35. (a) a = 5.28 m s (b) T = 45.27 N
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36. (a) a = 2.4 m s (b) F contact = 21.6 N
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37. (b) a = 0.7 m s
38. (a) f = 308.9 N (b) = 0.49
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39. a = 4.86 m s
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40. (b) a = 6.30 m s (c) T = 31.53 N
41. Newton’s third law of motion states that for every action there is a reaction of the same magnitude but
in opposite direction.
42. The pair of forces that acts when a hockey stick hits a ball based on Newton’s third law of motion are
action and reaction forces. They have the same magnitude but are at opposite directions to each other.
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43. (b) (ii) a = 5.04 m s
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