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Question

Block A of mass m2 is connected to one end of a light rope which passes over a light frictionless pulley as shown in figure.


A man of mass 2m climbs the other end of the rope with a relative acceleration of g2 with respect to rope. Find the acceleration (a0) of block A with respect to ground.

A
a0=g2
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B
a0=23g
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C
a0=32g
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D
a0=g
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Solution

The correct option is D a0=g
Let block A move up with acceleration a0 w.r.t the ground. Then rope which the man is holding will move downward with acceleration a0
Given : Acceleration of man w.r.t rope
amr=g2 (upwards)

We know amg=amr+arg
Taking upwards +ve,
amg=a=g2a0....(i)

From FBD of block:


Tmg2=ma02.......(ii)

From FBD of man:


T2mg=2ma.......(iii)

Substituting (i) in (iii):
T2mg=mg2ma0
T3mg=2ma0....(iv)

Subtracting (ii)(iv)
5mg2=5ma02
a0=g

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