A spring with spring constant $k=100\text{ N/m}$ is compressed by $x=0.1\text{ m}$. What is the potential energy stored in the spring?

Solution

1. Use the formula for potential energy $U=(1/2)k{x}^2$.
2. Substitute the given values $k=100\text{ N/m}$ and $x=0.1\text{ m}$.
3. Calculate $$\begin{array}{rl}U& =(1/2)\times 100\times (0.1{)}^2\\ & =50\times 0.01\\ & =0.5\text{ J}\end{array}$$.
4. Hence the answer is (A).