Elastic potential energy

Stretch or squash an ideal spring and the work you put in is stored: $U_s = \frac{1}{2}k(\Delta x)^2$, printed on all four AP equation sheets. $k$ is the spring constant in N/m, and $\Delta x$ is, in the CED's words, the distance the spring has been stretched or compressed from its equilibrium length.

The square settles two questions at once. Sign does not matter, so pulling a spring 4 cm out and pushing it 4 cm in store exactly the same energy, and $U_s$ can never come out negative. And the growth is quadratic: stretch it three times as far and it holds nine times the energy.

Measure $\Delta x$ from the spring's natural length. Not from the floor, not from wherever the block happens to be sitting. That single reference error is the usual reason a spring energy answer comes out wrong.

Keep the energy apart from the force. The spring force $\vec{F}_s = -k\Delta\vec{x}$ is linear in the stretch and is a vector; the stored energy is quadratic and is a scalar. A spring pulled twice as far pulls back twice as hard but holds four times as much energy.