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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

\[\begin {array} {1 1} (a)\;\frac{w}{4} & \quad (b)\;\frac{w}{2} \\ (c)\;\frac{3w}{4} & \quad (d)\;w \end {array}\]

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1 Answer

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$(c) \frac{3w}{4}$
answered Nov 7, 2013 by pady_1
 

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