Q. $K$ is the force constant of a spring. The work done in increasing its extension from $l_{1}$ to $l_{2}$ will be
A) $K(l_{2} - l_{1})$
B) $\frac{K}{2}(l_{2} + l_{1})$
C) $K(l_{2}^{2} - l_{1}^{2})$
D) $\frac{K}{2}(l_{2}^{2} - l_{1}^{2})$
Explanation
(d) At extension $l_{1}$, the stored energy $= \frac{1}{2}Kl_{1}^{2}$
At extension $l_{2}$, the stored energy $= \frac{1}{2}Kl_{2}^{2}$
Work done in increasing its extension from $l_{1}$ to $l_{2}$
$= \frac{1}{2}K(l_{2}^{2} - l_{1}^{2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.