Answer
$$11.25 \mathrm{J}$$
Work Step by Step
First, we determine the value of the spring constant as follows:
\begin{align*}
\int_{0}^{0.1} k x d x&=\left.\frac{1}{2} k x^{2}\right|_{0} ^{0.1}\\
&=0.005 k\\
&=5 \mathrm{J}
\end{align*}
Thus, $k = 1000$ N/m. Next, we calculate the work required to stretch the spring 15 cm beyond equilibrium:
\begin{align*}
\int_{0}^{0.15} 1000 x d x&=\left.500 x^{2}\right|_{0} ^{0.15}\\
&=11.25 \mathrm{J}
\end{align*}
