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Question
A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms–2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
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Solution 1
(a) Mass of the block, m = 15 kg
Coefficient of static friction, μ = 0.18
Acceleration of the trolley, a = 0.5 m/s2
As per Newton’s second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:
F = ma = 15 × 0.5 = 7.5 N
This force is acted in the direction of motion of the trolley.
Force of static friction between the block and the trolley:
f = μmg
= 0.18 × 15 × 10 = 27 N
The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest.
When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.
(b) An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.
Solution 2
(a) Force experienced by block, F = ma = 15 x 0.5 = 7.5 N Force of friction,Ff= p mg = 0.18 x 15 x 10 = 27 N. i.e., force experienced by block will be less than the friction.So the block will not move. It will remain stationary w.r.t. trolley for a stationary observer on ground.
(b) The observer moving with trolley has an accelerated motion i.e., he forms non-inertial frame in which Newton’s laws of motion are not applicable. The box will be at rest relative to the observer.
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