## College Physics (7th Edition)

$E_{1}=10J$ $E_{2}=20J$
$x_{1}=\frac{F}{k_{1}}=\frac{100}{500}=0.2m$ $x_{2}=\frac{F}{k_{2}}=\frac{100}{250}=0.4m$ Therefore, $E_{1}=\frac{k_{1}x_{1}^{2}}{2}=\frac{500\times 0.2^{2}}{2}=10J$ $E_{2}=\frac{k_{2}x_{2}^{2}}{2}=\frac{250\times 0.4^{2}}{2}=20J$