question_answer

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)  [Pb. PMT 2002]

A)             1 m/s2

B)             2.5 m/s2

C)             10 m/s2

D)               5 m/s2

Correct Answer: D

Solution :

                Net downward acceleration \[=\frac{\text{Weight-Friction force}}{\text{Mass}}\]             \[=\frac{(mg-\mu \ R)}{m}\]             \[=\frac{60\times 10-0.5\times 600}{60}\]             \[=\frac{300}{60}=5\ m/{{s}^{2}}\]