Question

A boy of mass m is sliding down a vertical pole by pressing it with a horizontal force $f$. If $\mu$ is the coefficient of friction between his palms and the pole, the acceleration with which he slides down will be equal to

• A. $g$
• B. $\frac{\mu f}{m}$
• C. $g+\frac{\mu f}{m}$
• D. $g-\frac{\mu f}{m}$

Answer 1

The correct option is D

$g-\frac{\mu f}{m}$

Here, normal reaction, $R=f$
Therefore, force of friction, $f_r=\mu R=\mu f$
Net downward force, $F=mg-f_r=mg-\mu f$
From Newton's second law of motion, acceleration, $a=\frac{F}{m}=\frac{mg-\mu f}{m}=g-\frac{\mu f}{m}$
Hence, the correct choice is (d).