    Question

# A man slides down on a telegraphic pole with an acceleration equal to ${\left(\frac{1}{4}\right)}^{th}$ of acceleration due to gravity. The frictional force between man and pole is equal to in terms of man's weight $w$

A

$\frac{w}{4}$

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B

$\frac{w}{2}$

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C

$\frac{3w}{4}$

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D

$w$

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Solution

## The correct option is C $\frac{3w}{4}$Step 1: Given data:Acceleration of the man, $a=\frac{g}{4}$ 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=w–f$ [$m$ is mass, $g$is acceleration due to gravity]$⇒f=w–ma$$⇒f=mg–m×\frac{g}{4}$ $\left[\because a=\frac{g}{4},w=mg\right]$Therefore, $f$$=\frac{3w}{4}$Hence, option (C) is correct.  Suggest Corrections  2      Similar questions  Explore more