JEE Main Physics question
A fireman of mass 60 kg slides down a pole. He is pressing the pole with a force of 600 N. The coefficient of friction between the hands and the pole is 0.5, with what acceleration will the fireman slide down (g = 10 m/s²)
Answer: (D)
Solution
Friction (upward) = μN = 0.5×600 = 300 N. Net downward force = mg − friction = 600 − 300 = 300 N. Acceleration = 300/60 = 5 m/s².