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? (\[g=10\,\,m{{s}^{-2}}\])

A)  \[1\,\,m{{s}^{-2}}\]            

B)  \[2.5\,\,m{{s}^{-2}}\]

C)  \[10\,\,m{{s}^{-2}}\]                      

D)  \[5\,\,m{{s}^{-2}}\]

Correct Answer: D

Solution :

 Net downward acceleration \[{{a}_{d}}=\frac{\text{Weight}-\text{friction force}}{\text{mass}}\] \[\Rightarrow \]            \[{{a}_{d}}=\frac{(mg-\mu R)}{m}\](Area R = normal force)             \[=\frac{60\times 10-0.5\times 600}{60}\] \[=\frac{300}{60}=5\,\,m{{s}^{-2}}\]