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Question: A fireman of mass 60kg slides down a pole. He is pressing the pole with a force of 600 N. The coeffi...

A fireman of mass 60kg 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)

A

1 m/s2

B

2.5 m/s2

C

10 m/s2

D

5 m/s2

Answer

5 m/s2

Explanation

Solution

Friction = μR = 0.5 × 600 = 300 N, Weight = 600 N

ma = W – F ⇒ a=W−Fm=600−30060a = \frac { W - F } { m } = \frac { 600 - 300 } { 60 }

∴ a=5 m/s2a = 5 \mathrm {~m} / \mathrm { s } ^ { 2 }