First, find the spring constant $k$ using Hooke's Law, $F = kx$.

$$k = \frac{F}{x_1}$$

The elastic potential energy $E_p$ stored in the spring is given by $E_p = \frac{1}{2}kx^2$. For a compression of $x_2$, the energy is:

$$E_p = \frac{1}{2}k x_2^2 = \frac{1}{2}\left(\frac{F}{x_1}\right)x_2^2$$

Substituting the values ($x_1=0.04$ m, $x_2=0.10$ m):

$$E_p = \frac{1}{2}\left(\frac{120 \text{ N}}{0.04 \text{ m}}\right)(0.10 \text{ m})^2 = 15 \text{ J}$$