College Physics (4th Edition)

The elastic potential energy stored in half a spring is $20.0~J$ when it is compressed a distance of $0.20~m$
Let $k$ be the spring constant of the original spring. When a force of $F$ compresses the spring, the spring compresses a distance of $\frac{F}{k}$. Before the spring is cut, each half of the spring pushes with a force of $F$ but only compresses a distance of $\frac{F}{2k}$, that is, half the total distance. We can find the spring constant $k'$ of each half of the spring after the original spring is cut: $k' = \frac{F}{\frac{F}{2k}} = \frac{2Fk}{F} = 2k$ We can find an expression for the elastic potential energy stored in half a spring when it is compressed a distance $x$: $U_s = \frac{1}{2}k'x^2$ $U_s = \frac{1}{2}(2k)x^2$ $U_s = 2\times \frac{1}{2}kx^2$ Since the elastic potential energy stored in half the spring is double the elastic potential energy for the original spring, the elastic potential energy stored in half a spring when it is compressed a distance of $0.20~m$ is $(2)(10.0~J)$ which is $20.0~J$.