PE = ½kx² — Elastic Potential Energy of a Spring

Elastic potential energy stored in a stretched or compressed spring — with a worked example.

PE = \frac{1}{2}kx^2

Spring potential energy equals one-half times the spring constant times the stretch squared.

What it means

A stretched or compressed spring stores energy proportional to the square of the displacement. The stiffness $k$ sets the scale: a stiff spring stores far more energy for the same stretch. Release it and this energy becomes kinetic.

Variables

Symbol	Name	Unit	Meaning
$PE$	Elastic potential energy	J	Energy stored in the deformed spring
$k$	Spring constant	N/m	Spring stiffness
$x$	Displacement	m	Stretch or compression from natural length

Assumptions

Spring obeys Hooke's law (F = kx) over the stretch used
x measured from the spring's relaxed length
No energy lost to heat or sound

Derivation

Work stored in the spring is $\int_0^x kx\,dx$ . Since the force grows linearly from 0 to $kx$ , the average force is $kx/2$ , giving work $\tfrac{1}{2}kx^2$ — stored as potential energy.

Worked example

Problem

A spring with stiffness 200 N/m is compressed 0.15 m. How much elastic potential energy is stored?

Given:

k = 200 N/m
x = 0.15 m

Write the formula: $PE = \tfrac{1}{2}kx^2$
Substitute: $PE = 0.5 \times 200 \times 0.15^2$
Calculate: $PE = 0.5 \times 200 \times 0.0225 = 2.25\ \mathrm{J}$

2.25 J

Compressing the spring stores 2.25 J — enough to launch a small toy. Because energy scales with x², doubling the compression quadruples the stored energy.