Q. 35

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 m s-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.


Answer :

Given:

Mass of block, M = 15 Kg


Coefficient of static friction, � = 0.18


Acceleration of the trolley, a = 0.5 ms-2


Time for which the trolley accelerates, t = 20 s


To check whether the block will slide or not, let us compare the force due to trolley and


trolley and Block.


Using the Newton’s second law of motion,


Force, F on the block due to accelerating trolley is given by,


F = M× a


F = 15 Kg× 0.5 ms-2


F = 7.5 N


Frictional force, fs


fs = �× M× g


fs = 0.18× 15Kg × 10 ms-2


fs = 27 N


F<fs


The force of friction between the block and the trolley is greater than the applied force by trolley. Hence for an observer on ground the block will seem to be moving at the same velocity as that of trolley.


b) For an observer moving with the trolley, the trolley and the box will seem to be at rest.


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