RGS Prelims 1992 P2 Q1
An object of mass 5 kg is moving with a constant velocity of 10 m s-1. A resultant force of 4 N is then applied to it in the opposite direction to its motion.
(a) If the force acts for 5 s and is then removed, what is the subsequent motion (velocity) of the object in the next 5 s?
(b) Draw the velocity-time graph showing the motion of the object in the first 10 s.
(c) Calculate the distance moved by the object in the first 5 s when the force is applied.
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Answer:
(a) Using F = ma, where F = -4 and m = 5
a = -4/5 = -0.8 s-2
After 5 s, object would have slowed down by 5 * 0.8 = 4 ms-1, which means after 5 s, the object would be moving at 6 ms-1.
Thus, the subsequent motion of the object in the next 5 s is that it will be moving at a constant velocity of 6 ms-1.
(b)
(c) There are two methods of solving. Note that during exams, it is good to use a 2nd method to verify your answer.
Method 1: Area under graph
Distance moved = ½ * (10 + 6) * 5 = 40 m
Method 2: Using kinematics equation
Distance moved = ut + ½at²
= (10)(5) + ½(-0.8)(5)² ===> note that acceleration is negative because it is a deceleration
= 50 - 10
= 40 m
