JEE Main Physics question
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, in terms of the man's weight w,
Answer: (C)
Solution
mg − f = mg/4 ⟹ f = mg − mg/4 = 3mg/4 = 3w/4.