Q. 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) $5\ m/s^{2}$
Explanation
(d) Net downward acceleration $= \frac{Weight - Friction\ force}{Mass}$
$= \frac{(mg - \mu R)}{m}$
$= \frac{60 \times 10 - 0.5 \times 600}{60}$
$= \frac{300}{60} = 5\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.