31. 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 a) \( \frac{w}{4} \) b) \( \frac{w}{2} \) c) \( \frac{3 w}{4} \) d) \( W \)

Question

31. 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 a) \( \frac{w}{4} \) b) \( \frac{w}{2} \) c) \( \frac{3 w}{4} \) d) \( W \)

Answer 1

Correct option is c) \(\frac{3w}{4}\)

f = mg - ma

= m(g - g/4)

= 3/4 mg

= 3/4 W