﻿ 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. from Physics Laws of Motion Class 11 Manipur Board

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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.

Given,

Mass of the block, m = 15 kg

Coefficient of static friction, μ = 0.18

Acceleration of the trolley, a = 0.5 m/s

According to Newton’s second law of motion,

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 = μm

= 0.18 × 15 × 10

= 27 N

The applied external force is less than the force of static friction between the block and the trolley. Therefore, the block will appear to be at rest, for an observer on the ground.

When the trolley moves with uniform velocity, only the force of friction will act in this situation and there will be no applied external force.

(b) The person who is moving with the trolley has some acceleration.

The frictional force is acting backwards on the trolley and is opposed by a pseudo force of the same magnitude. But, this force is acting in the backward direction. Therefore, for an observer moving with the trolley, the trolley will appear to be at rest.
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