Elastic Potential Energy

What elastic potential energy is

When a spring is compressed or stretched, it stores energy. Release it, and this energy converts to kinetic energy. This stored energy is called elastic potential energy:

$U = \frac{1}{2}kx^2$

where $k$ is the spring constant and $x$ is the displacement from the natural (equilibrium) length.

Derivation

Elastic PE is defined as the negative of work done by the spring:

$U = -W_{spring}$

Work done by spring force $F = -kx$ as body moves from $0$ to $x$:

$W_{spring} = \int_0^x (-kx') \, dx' = -\frac{1}{2}kx^2$

Therefore:

$U = -W_{spring} = -\left(-\frac{1}{2}kx^2\right) = \frac{1}{2}kx^2$

$\boxed{U = \frac{1}{2}kx^2}$

Key properties

Always positive: $U = \frac{1}{2}kx^2 \geq 0$. Both compression ($x < 0$) and extension ($x > 0$) give positive PE — the spring stores energy in both cases.

Zero at natural length: $U = 0$ when $x = 0$. No energy stored when spring is at its natural length.

Quadratic in displacement: Doubling the compression quadruples the stored energy.

Spring PE vs gravitational PE

Energy stored in a spring — the parabola

Plot $U$ vs $x$: it is a parabola opening upward with minimum at $x = 0$. The system naturally tends toward minimum energy — the spring's equilibrium position.

Spring-mass system energy conservation

A mass $m$ on a spring, released from extension $x_0$:

At $x = x_0$ (released from rest): $KE = 0$, $U = \frac{1}{2}kx_0^2$

At $x = 0$ (natural length): $U = 0$, $KE = \frac{1}{2}kx_0^2$

Speed at natural length: $v = x_0\sqrt{\frac{k}{m}}$

Total energy is conserved throughout: $\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kx_0^2$