Spring Energy⚠ unverified
Mechanical / Springs · Compute the elastic energy stored in a spring
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| F | F | N | 1.0 | Applied force |
| delta | δ | m | 1.0 | Resulting deflection |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | U | J | Stored elastic energy, in joules (J) |
The science & history
Understanding the Parameters
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Applied force $F$ — the force at the final deflected state. It is not the force that does all the work: the spring resisted with less force earlier in the stroke, which is why the energy is $\tfrac12 F\delta$ and not $F\delta$.
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Deflection $\delta$ — the distance moved. Energy scales linearly with both $F$ and $\delta$, and since they are themselves proportional ($F = k\delta$), energy actually scales with the square of either: $U = \tfrac12 k\delta^2 = \tfrac12 F^2/k$. Doubling the deflection quadruples the stored energy.
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The $\tfrac12$ — the fraction that distinguishes gradual (elastic) loading from a constant force. A weight dropped onto a spring delivers $F\delta$ of work but the spring stores only $\tfrac12 F\delta$ at equal deflection — the difference drives the overshoot/oscillation of impact loading.
Derivation (Approaching a Proof)
Work done on the spring equals the integral of force over the deflection path. For a linear spring the force rises in proportion to deflection, $f(x) = k x$, from $0$ to the final $\delta$:
$$U = \int_0^{\delta} f(x)\,\mathrm{d}x = \int_0^{\delta} k x\,\mathrm{d}x = \tfrac{1}{2} k \delta^2.$$
Substituting the linear relation $F = k\delta$ (so $k\delta = F$) gives the force–deflection form:
$$U = \tfrac{1}{2} (k\delta)\,\delta = \tfrac{1}{2} F \delta.$$
Geometrically, $U$ is the area under the force–deflection line — a triangle of base $\delta$ and height $F$, hence $\tfrac12 F\delta$. Energy is conserved and fully recoverable for an ideal elastic spring; real springs lose a little to hysteresis. This is the same $\tfrac12$(generalised force)(generalised displacement) that gives $\tfrac12 CV^2$ for a capacitor and $\tfrac12 LI^2$ for an inductor.
Dimensional check. $[U] = \text{N}\cdot\text{m} = \text{J}$. ✓
History and Development
The result is a direct consequence of Hooke's law (Robert Hooke, 1678, ut tensio, sic vis — "as the extension, so the force") integrated over the deflection. Spring energy storage underlies the entire history of portable mechanical power — spring-driven clocks and watches from the 15th century onward, and modern applications from mechanical watches to regenerative valve springs. The quadratic energy law $U = \tfrac12 k\delta^2$ is the mechanical member of the family of quadratic energy-storage laws across physics.
Related Concepts: Helical Spring Rate, Helical Spring Deflection, Strain Energy Density, Elastic Potential Energy, Spring Natural Frequency
Notes: For an ideal linear spring. A preloaded spring stores $\tfrac12 k(\delta_2^2 - \delta_1^2)$ between two deflections. Nonlinear springs (e.g. Belleville, progressive) require integrating the actual $F(\delta)$ curve.