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Science

Spring Potential Energy Calculator

Elastic potential energy is the energy stored in an ideal Hookean spring when it is stretched or compressed. Displace the spring by x from its natural length and it stores U = ½ · k · x², where k is the spring constant (N/m). This tool accepts k in N/m, N/cm, N/mm or lbf/in and x in m, cm, mm or in, normalises every input to SI, and shows U in joules together with kilojoules, kilocalories, foot-pounds and watt-hours plus the matching spring force F = k·|x|.

Elastic potential energy U

Spring force at this displacement, F = k·x

Other energy units

Kilojoules
Kilocalories
Foot-pounds force
Watt-hours

1 Wh = 3600 J — handy for comparing against household appliance energy use.

Formula

U = ½ · k · x² F = k · |x| (k in N/m, x in m, result in J = N·m)

Elastic energy scales with the square of displacement — double x and U quadruples. Inputs are normalised to N/m and m before the formula is applied; the sign of x records direction (positive = stretched, negative = compressed) but force and energy are taken as magnitudes.

Formula

U = ½ · k · x² // elastic potential energy (joules, with k in N/m and x in m) F = k · |x| // matching spring-force magnitude (newtons)

Frequently asked

How is elastic potential energy different from gravitational potential energy?

Both are stored energies but with very different mechanisms. Gravitational potential energy U_g = m·g·h depends linearly on height and always decreases downward. Elastic potential energy U_s = ½·k·x² depends on the square of displacement and increases whether the spring is stretched or compressed — both directions store energy. Everyday example: lifting a book from the floor to a desk increases gravitational PE; placing the same book on a kitchen scale and compressing its spring by 2 cm increases elastic PE. In a slingshot, bow, or vehicle suspension, the elastic energy released converts to kinetic energy or heat.

Why is there a factor of ½ in U = ½·k·x²?

Because stretching the spring from 0 to x requires a force that grows linearly from 0 to kx — you are not pulling against the peak force kx the whole way. The average force is only ½·kx, and the work done is average force × distance = ½·kx · x = ½·k·x². From calculus: U = ∫₀ˣ k·s ds = ½·k·x². Geometrically the F-x graph is a triangle and the area (= work) is ½ · base · height. The same ½ appears in the capacitor (U = ½·C·V²) and inductor (U = ½·L·I²) storage formulas for exactly the same linear-growth reason.

When does U = ½·k·x² stop being accurate?

Whenever the deformation goes past the proportional limit and Hooke's law no longer holds. Every real spring has an elastic limit (beyond which it deforms permanently) and a fracture point. Rubber bands, rubber, biological tissue, conical springs, progressive coil springs and Belleville washers are non-linear from the start and need the more general U = ∫ F(x) dx. Under very fast loading (impact, high-frequency oscillation) internal damping (hysteresis) dissipates part of the work as heat, so the energy actually stored is a few percent below ½·k·x². Engineers typically restrict working strain to 50–70% of the elastic range to keep behaviour linear and avoid fatigue.

How do joules (J), calories and foot-pounds force relate to each other?

1 joule is the SI energy unit — the work of 1 newton over 1 metre, or 1 watt for 1 second. Exact conversions (NIST SP 811, 2008): 1 thermochemical kilocalorie = 4184 J (1 cal = 4.184 J); 1 foot-pound force = 1.3558179483314004 J; 1 watt-hour = 3600 J. So stretching a 500 N/m spring by 10 cm stores U = 0.5·500·0.10² = 2.5 J — about 0.000597 kcal, 1.844 ft·lbf or 0.000694 Wh. The tool prints all four units side by side so you can compare against food calories, mechanical engineering specs or household-appliance energy use.

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