Answer
-0.156 J
Work Step by Step
Given:
$x_{i}=7.60\,mm=7.60\times10^{-3}\,m$,
$x_{f}=(7.60+7.60)\,mm=15.2\times10^{-3}\,m$ and
spring constant $k=18.0\,N/cm=\frac{18.0\,N}{10^{-2}m}$
Recall:
Work done by the spring force $W_{s}=\frac{1}{2}kx_{i}^{2}-\frac{1}{2}kx_{f}^{2}$
Result:
$W_{s}=\frac{1}{2}\times\frac{18.0\,N}{10^{-2}m}\times((7.60\times10^{-3}\,m)^{2}-(15.2\times10^{-3}\,m)^{2})$
$=-0.156\,J$
