Question

Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass is
(a) $\frac{1}{2}k{x}^2$
(b) $-\frac{1}{2}k{x}^2$
(c) $\frac{1}{4}k{x}^2$
(d) $-\frac{1}{4}k{x}^2$

Answer 1

(d) $-\frac{1}{4}k{x}^2$

The work done by the spring on both the masses is equal to the negative of the increase in the elastic potential energy of the spring.

The elastic potential energy of the spring is given by $E_p=\frac{1}{2}k{x}^2$.

Work done by the spring on both the masses = $-\frac{1}{2}k{x}^2$
∴ Work done by the spring on each mass = $\frac{1}{2}\left(-\frac{1}{2}k{x}^2\right)=-\frac{1}{4}k{x}^2$