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A man slides down on a telegraphic pole ...

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

A

`w/4`

B

`w/2`

C

`(3w)/4`

D

w

Text Solution

Verified by Experts

The correct Answer is:
C
Acceleration =` ¼` Acceleration due to gravity
`mg-F = (mg)/4`
`implies=(3mg)/4`
` therefore w=mg`
` therefore F=(3w)/4`
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