Question

One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is $\frac{1}{2}k{x}^2$. The possible cases are
(a) at spring was initially compressed by a distance x and was finally in its natural length
(b) it was initially stretched by a distance x and and finally was in its natural length
(c) it was initially in its natural length and finally in a compressed position
(d) it was initially in its natural length and finally in a stretched position.

Answer 1

(a) at spring was initially compressed by a distance x and was finally in its natural length
(b) it was initially stretched by a distance x and and finally was in its natural length

For an elastic spring, the work done is equal to the negative of the change in its potential energy.

When the spring was initially compressed or stretched by a distance x, its potential energy is given by
${\left(P.E.\right)}_i=\frac{1}{2}k{x}^2$.

When it finally comes to its natural length, its potential energy is given by
${\left(P.E.\right)}_f=0$.

∴ Work done = $-\left[{\left(P.E.\right)}_f-{\left(P.E.\right)}_i\right]=-\left[0-\frac{1}{2}k{x}^2\right]=\frac{1}{2}k{x}^2$