Q. A child weighing $25\ kg$ slides down a rope hanging from the branch of a tall tree. If the force of friction acting against him is $2\ N$, what is the acceleration of the child (Take $g = 9.8\ m/s^{2}$)
A) $22.5\ m/s^{2}$
B) $8\ m/s^{2}$
C) $5\ m/s^{2}$
D) $9.72\ m/s^{2}$
Explanation
(d) $Net\ download\ force\ = \ Weight\ –\ Friction$
$\therefore ma = 25 \times 9.8 - 2 \Rightarrow a = \frac{25 \times 9.8 - 2}{25} = 9.72\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.