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

The correct answer/option is -kx²/4.

Let us analyze the problem step-by-step:

Suppose we have a spring of spring constant $k$ aligned along the horizontal axis, with two equal masses attached to its two ends. Let the natural length of the spring be $L_0$.

The masses are pulled out symmetrically. Symmetrical stretching means that both ends are displaced by equal distances in opposite directions relative to the center of the spring (which remains stationary).

If the total elongation of the spring is $x$, then each end is stretched by a distance of:
$d=\frac{x}{2}$
away from the center of the spring.

The total potential energy stored in the spring when it is stretched by a total length $x$ from its natural length is given by:
$U=\frac{1}{2}k{x}^2$

The total work done by the spring on both masses during this stretching process is equal to the negative change in the potential energy of the spring (since the initial potential energy was zero when the spring was at its natural length):
$W_{\text{total}}=-U=-\frac{1}{2}k{x}^2$

Because the masses are equal and the stretching is perfectly symmetrical, the spring forces acting on each mass are equal in magnitude at all instants, and each mass undergoes the same displacement magnitude. Thus, the total work done by the spring is shared equally between the two masses.

Therefore, the work done by the spring on each mass ($W_{\text{each}}$) is half of the total work done by the spring:
$W_{\text{each}}=\frac{W_{\text{total}}}{2}=\frac{-\frac{1}{2}k{x}^2}{2}=-\frac{1}{4}k{x}^2$

This can also be written as:
$-{\text{kx}}^2/4$
The negative sign indicates that the restoring force exerted by the spring on each mass points inward (towards the center), which is opposite to the outward displacement of the masses.