Answer
$\Delta U_e = 1.0~J$
Work Step by Step
We can find the change in elastic potential energy:
$\Delta U_e = \frac{1}{2}ky_2^2-\frac{1}{2}ky_1^2$
$\Delta U_e = \frac{1}{2}ky_2^2-0$
$\Delta U_e = \frac{1}{2}(200~N/m)(-0.10~m)^2$
$\Delta U_e = 1.0~J$
