The potential energy of a spring stretched by a distance $x$ is given by $U = \frac{1}{2}kx^2$ . Given that the potential energy for a stretch $S$ is $10 \text{ J}$: $U_1 = \frac{1}{2}kS^2 = 10 \text{ J}$ When the spring is stretched through an additional distance $S$, the total stretch becomes $S + S = 2S$. The new potential energy of the spring is: $U_2 = \frac{1}{2}k(2S)^2 = 4 \times \left(\frac{1}{2}kS^2\right) = 4 \times 10 = 40 \text{ J}$ The work done by the external force to stretch the spring through the additional distance is equal to the change in its elastic potential energy ($W = \Delta U$): $W = U_2 - U_1 = 40 \text{ J} - 10 \text{ J} = 30 \text{ J}$.