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🪝 Hooke's Law Calculator (F = kx)

Find the restoring force of a spring (F = kx), or solve for the spring constant or extension.

Force F

10 N

Hooke’s law: F = k·x (restoring force = spring constant × extension). Elastic PE = ½·k·x². 🔒 Computed in your browser.

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How the hooke's law calculator (f = kx) works

Hooke’s law says a spring’s restoring force is proportional to its extension: F = kx, where k is the spring constant (stiffness) and x the extension or compression. Enter any two and the tool solves for the third. The energy stored is ½kx². The minus sign in the full form F = −kx signals that the force opposes the displacement, a restoring force pulling the spring back to its natural length; this tool works with magnitudes.

Valid within the elastic limit, stretch a spring too far and it deforms permanently and Hooke’s law breaks down. A stiffer spring has a larger k.

Frequently asked questions

What is Hooke’s law?

The force a spring exerts is proportional to how far it’s stretched or compressed: F = kx. Double the extension, double the force, up to the elastic limit.

How do I find the spring constant?

Rearrange to k = F/x. A spring needing 10 N to stretch 0.05 m has k = 200 N/m. Enter force and extension to get k.

What is the spring constant’s unit?

Newtons per metre (N/m). A larger k means a stiffer spring that resists stretching more.

How much energy is stored in a stretched spring?

Elastic potential energy = ½kx². For k = 200 N/m stretched 0.05 m, that’s ½ × 200 × 0.05² = 0.25 J.

When does Hooke’s law stop working?

Beyond the elastic limit, where the material deforms permanently. Then force is no longer proportional to extension.

Worked example: a loaded spring

Hang a 2 kg mass on a spring of k = 50 N/m. The stretching force is mg = 19.62 N, so the extension is x = F/k = 19.62 / 50 ≈ 0.39 m, and the stored energy is ½kx² ≈ 3.85 J.

How do springs combine?

Springs in parallel share the load, so their stiffnesses add (k_total = k₁ + k₂). In series they stretch more, so the reciprocals add: 1/k_total = 1/k₁ + 1/k₂, the reverse of how resistors behave.

What is the elastic limit?

The greatest extension for which a spring springs back to its natural length. Below it the spring is elastic and obeys F = kx; stretch it past the elastic limit and it deforms permanently, so the law no longer holds.

Why is the stored energy ½kx² and not kx²?

Because the force builds up from zero to kx as the spring stretches, so the average force is ½kx. Work equals average force times distance, giving ½kx·x = ½kx², the triangular area under the force, extension line.

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