A man slides down on a telegraphic pole with an acceleration $a = g / 4$ . The frictional force between the man and the pole is equal to (in terms of man's weight $W$ )

\frac{W}{4}

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\frac{W}{2}

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\frac{3 W}{4}

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Solution

The correct option is B $\frac{3 W}{4}$
$f = m g - m a$
$= m (g - g / 4)$
$= \frac{3}{4} m g$
$= \frac{3}{4} W$

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Multiply (w + 1) and 4w:

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A man slides down on a telegraphic pole with an acceleration a=g/4. The frictional force between the man and the pole is equal to (in terms of man's weight W)

The correct option is B 3W4f=mg−ma =m(g−g/4) =34mg =34W

A man slides down on a telegraphic pole with an acceleration $a = g / 4$ . The frictional force between the man and the pole is equal to (in terms of man's weight $W$ )

The correct option is B $\frac{3 W}{4}$
$f = m g - m a$
$= m (g - g / 4)$
$= \frac{3}{4} m g$
$= \frac{3}{4} W$