Q.
A vertical spring with force constant $K$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is
A) $mg + (h + d) + \frac{1}{2}Kd^{2}$
B) $mg(h + d) - \frac{1}{2}Kd^{2}$
C) $mg(h - d) - \frac{1}{2}Kd^{2}$
D) $mg(h - d) + \frac{1}{2}Kd^{2}$
Explanation
(b) Gravitational potential energy of ball gets converted into elastic potential energy of the spring $mg(h + d) = \frac{1}{2}Kd^{2}$
Net work done $= mg(h + d) - \frac{1}{2}Kd^{2} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.