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Question -

A block of mass 15 kg is placed on a longtrolley. The coefficient of static friction between the block and the trolleyis 0.18. The trolley accelerates from rest with 0.5 m s–2 for20 s and then moves with uniform velocity. Discuss the motion of the block asviewed by (a) a stationary observer on the ground, (b) an observer moving withthe trolley.



Answer -

(a) Mass of the block, m = 15kg

Coefficient of static friction, μ =0.18

Acceleration of the trolley, a =0.5 m/s2

As per Newton’s second law of motion, theforce (F) on the block caused by the motion of the trolley is given bythe relation:

F = ma = 15 × 0.5 = 7.5N

Thisforce is acted in the direction of motion of the trolley.

Forceof static friction between the block and the trolley:

f = μmg

=0.18 × 15 × 10 = 27 N

Theforce of static friction between the block and the trolley is greater than theapplied external force. Hence, for an observer on the ground, the block willappear to be at rest.

Whenthe trolley moves with uniform velocity there will be no applied externalforce. Only the force of friction will act on the block in this situation.

(b) An observer, moving withthe trolley, has some acceleration. This is the case of non-inertial frame ofreference. The frictional force, acting on the trolley backward, is opposed bya pseudo force of the same magnitude. However, this force acts in the oppositedirection. Thus, the trolley will appear to be at rest for the observer movingwith the trolley.

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