Solution to 2008 Problem 28

The potential energy stored in a spring with spring constant k that is stretched a distance d from its equilibrium position is

\begin{align*}U = \frac{1}{2}k d^2\end{align*}
Let k' denote the spring constant of the second spring. Then, from the problem statement,

\begin{align*}2 \cdot \left(\frac{1}{2}k d^2\right)  = \frac{1}{2} k' \left(\frac{d}{2}\right)^2\end{align*}
Solving for k' gives

\begin{align*}k' = \boxed{8 k}\end{align*}
Therefore, answer (D) is correct.

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