Setup: Spring with natural length $L_0$, spring constant k. Extension/compression: x = deformation from natural length.

Work done by spring from $x_1$ to $x_2$: $W_{spring}$ = $\frac{1}{2}$k*$x_1^2$ - $\frac{1}{2}$k*$x_2^2$

Note: This is (initial PE - final PE), which equals -$delta_{PE}$.

Special cases:

1. Stretching from 0 to x: W = -\frac{1}{2}$$kx^2 (spring does negative work)
2. Releasing from x to 0: W = +\frac{1}{2}$$kx^2 (spring does positive work)
3. Stretching from x to 2x: W = -$\frac{1}{2}$k(4x^{2-x}^2) = -\frac{3}{2}$$kx^2

Common JEE trap: Work to stretch from natural length to x is \frac{1}{2}$$kx^2. But work to stretch from x to 2x is NOT another \frac{1}{2}$$kx^2 — it is \frac{3}{2}$$kx^2 (three times as much!). Spring force increases with extension.

Maximum compression/extension: When all KE is converted to spring PE: \frac{1}{2}$$mv^2 = \frac{1}{2}$$kx_{max}^2, $x_{max}$ = v*sqrt$\frac{m}{k}$.