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**Hint:**The potential energy when the spring is stretched through a distance $S$ is $10$ joules. We need to find the work done when the spring is stretched for additional distance $S$ . Calculate the difference in the final and initial potential energy. The difference in the amount of the energy must be equal to the work done.

**Complete step by step solution:**

We are given the energy when a spring is stretched through a distance $S$ . We need to calculate the work done when the same spring is stretched by additional same distance. As the spring is the same, the spring constant will have the same value.

The initial potential energy $U$ of the spring is given as:

$U = \dfrac{1}{2}K{S^2}$

Here, $K$ is the spring constant

$S$ is the initial distance by which the spring is stretched.

We are given that $U = 10\,J$ , substituting this value in above equation, we get

$10 = \dfrac{1}{2}K{S^2}$--equation $1$

Finally, the spring is stretched by a distance of $2S$ , thus the final potential energy ${U_f}$ will be;

${U_f} = \dfrac{1}{2}K{(2S)^2}$

$ \Rightarrow {U_f} = 4 \times \dfrac{1}{2} \times K{S^2}$

But from equation $1$ , we have $\dfrac{1}{2}K{s^2} = 10$ , therefore we have

$ \Rightarrow {U_f} = 4 \times (10)$

$ \Rightarrow {U_f} = 40\,J$

As the final potential energy is $40\,J$ and the initial potential energy is $10\,J$ . The change in potential energy will be the work done. Therefore, the work done $W$ is given as:

$W = 40\,J - 10\,J$

$ \Rightarrow W = 30\,J$

**Thus, option A is the correct option.**

**Note:**To change the distance of the spring from $S$ to $2S$ , work is done on the spring. Remember this work done is responsible for the change in the potential energy of the spring. As the same spring is used in both the cases, the spring constant will remain the same.

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