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Question

A man slides down on a telegraphic pole with an acceleration equal to 14th of acceleration due to gravity. The frictional force between man and pole is equal to in terms of man's weight w


A

w4

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B

w2

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C

3w4

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D

w

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Solution

The correct option is C

3w4


Step 1: Given data:

Acceleration of the man, a=g4

Weight of the man, w=mg

Step 2: Balancing forces acting on man:

Friction force f will act opposite to the weight of a man, w

ma=wf [m is mass, gis acceleration due to gravity]

f=wma

f=mgm×g4 [a=g4,w=mg]

Therefore, f=3w4

Hence, option (C) is correct.


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