Question

A man of mass $65\, kg$ slides down a telegraphic pole with an acceleration equal to $\frac{3}{4}$ of acceleration due to gravity. The frictional force between the man and pole is_________ $N$. (Take $g =10\, m / s ^{2}$ )

Answer 1

Man is sliding down the telegraphic pole with acceleration $\frac{3 g }{4}$.
$\therefore mg - F =\frac{3 mg }{4}$
$\therefore F = mg -\frac{3 mg }{4}$
$\therefore F =\frac{ mg }{4}$
$\therefore F =\frac{65 \times 10}{4}$
$\therefore F =162.5\, N$