For a fixed spring constant, elastic energy follows a parabola. The marker shows the current deformation; the curve extends to twice that value.
At twice the deformation, the spring stores four times the energy.
The energy equals the area under the linear force–deformation graph: force rises from 0 to kx, so the triangular area is ½ × x × kx. In SI units, k is in newtons per metre, x is in metres, and U is in joules.
| Solve for | Rearranged formula | SI result |
|---|---|---|
| Energy U | U = ½kx² | joules (J) |
| Spring constant k | k = 2U/x² | newtons per metre (N/m) |
| Deformation magnitude x | x = √(2U/k) | metres (m) |
| Force magnitude F | F = kx | newtons (N) |
Reference point: this page assigns zero elastic energy to the undeformed spring. Adding a different arbitrary potential-energy constant changes absolute U values but not energy differences or forces.
Given: k = 250 N/m and x = 8 cm = 0.08 m.
Work: U = ½(250)(0.08)² = 0.8 J.
Given: U = 2 J and x = 10 cm = 0.1 m.
Work: k = 2(2)/(0.1)² = 400 N/m.
Given: U = 9 J and k = 800 N/m.
Work: x = √(18/800) = 0.15 m.
This calculator models a passive, ideal spring with constant stiffness k. It assumes the deformation is within the elastic, approximately linear range and that energy lost to hysteresis, friction, damping, heat, sound, and permanent deformation is negligible. Use measured force–displacement data or manufacturer limits for real springs, elastomers, gas springs, composite structures, and safety-critical designs.
The deformation field is a magnitude. Compression and extension of equal magnitude produce the same result in this ideal model.
For an ideal linear spring, elastic potential energy is U = ½kx², where k is the spring constant and x is the deformation from the unstretched position.
No. Because x is squared, equal-magnitude compression and extension store the same nonnegative energy when the undeformed spring is the zero-energy reference.
Using k in N/m and x in m gives U in joules. The calculator converts the other listed units to SI before applying the formula.
Energy depends on x². Replacing x with 2x multiplies the energy by 2², or four.
It is inaccurate when stiffness changes substantially with deformation or energy is not fully recoverable—for example near yielding, coil bind, large geometric deformation, hysteresis, or friction.
For a standalone ideal spring, x is measured from its undeformed natural length. For a vertical oscillator, a gravity-shifted equilibrium can be used if elastic and gravitational potential energy are combined consistently.
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