A stretched spring can launch a toy or start a clock, so it stores energy. How much? It equals the work done to deform it, which is the area under the force-stretch graph.
Key idea
A spring stretched or compressed by $x$ from its natural length stores elastic potential energy $U_s = \tfrac12 kx^2$.
Why it works
The force needed grows from $0$ to $kx$ as you stretch the spring (Hooke's law), so the $F$–$x$ graph is a straight line through the origin. The work is the triangle under it: $\tfrac12 \times x \times kx = \tfrac12 kx^2$. The $\tfrac12$ comes from the triangle; the square comes from the force and the distance both growing with $x$.
Worked examples
Worked example: Energy stored
A spring with $k = 250\ \text{N/m}$ is compressed $12\ \text{cm}$. How much energy does it store?
- $x = 0.12\ \text{m}$.
- $U_s = \tfrac12(250)(0.12)^2 = 1.8\ \text{J}$.
Answer: $1.8\ \text{J}$
Worked example: Stretch from energy
How far must a spring with $k = 500\ \text{N/m}$ be stretched to store $4\ \text{J}$?
- $4 = \tfrac12(500)x^2$, so $x^2 = 0.016$.
- $x \approx 0.126\ \text{m} = 12.6\ \text{cm}$.
Answer: about $12.6\ \text{cm}$
Worked example: Adding stretch
A spring with $k = 300\ \text{N/m}$ is already stretched $5\ \text{cm}$. How much more energy does it take to stretch it to $10\ \text{cm}$?
- $$\Delta U = \tfrac12(300)(0.10^2 - 0.05^2) = 150(0.0075) = 1.125\ \text{J}.$$
- Not $\tfrac12(300)(0.05)^2 = 0.375\ \text{J}$: the second $5\ \text{cm}$ is harder than the first.
Answer: $1.125\ \text{J}$
Because the energy grows as $x^2$, the last few centimeters of a stretch store the most energy: going from $0.1$ to $0.2\ \text{m}$ above adds $6\ \text{J}$, three times as much as the first $0.1\ \text{m}$.
Common mistake
Writing $kx^2$ without the $\tfrac12$, or using the force $kx$ as if it were the energy. Check the units: $kx$ is in N, a force; $\tfrac12 kx^2$ is in N·m = J, an energy. And doubling the stretch should multiply the energy by four, not two.