A man slides down a telegraphic pole with an | vedclass

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(C) Let the mass of the man be $m$. The weight of the man is $w = mg$.The forces acting on the man are his weight $mg$ acting downwards and the fricti

Vedclass · Accepted Answer

$\frac{3w}{4}$

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(C) Let the mass of the man be $m$. The weight of the man is $w = mg$.The forces acting on the man are his weight $mg$ acting downwards and the frictional force $F$ acting upwards.According to Newton's second law of motion,the net force acting on the man is equal to the product of his mass and acceleration.Since the man is sliding down with an acceleration $a = g/4$,we have:$mg - F = ma$Substituting $a = g/4$:$mg - F = m(g/4)$$mg - F = mg/4$$F = mg - mg/4$$F = \frac{3mg}{4}$Since $w = mg$,the frictional force is $F = \frac{3w}{4}$.

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

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