BYTETOOLS

Hooke's Law Calculator

Solve F = kx for force, spring constant or extension, get the stored energy from U = ½kx², and combine springs in series or parallel.

20 N
Restoring force
100 N/m
Spring constant
0.2 m
Extension
2 J
Stored energy

F = k · x

  • F = 100 N/m × 0.2 m = 20 N
  • Elastic potential energy U = ½·k·x² = 0.5 × 100 N/m × (0.2 m)² = 2 J (2 J)
  • Work to stretch from 0 m to 0.2 m = ½k(x₂² − x₁²) = 2 J
  • The minus sign in F = −kx says the spring pulls back towards its rest position; the magnitude is what you read above.

Work done between two extensions

Because the force grows as you stretch, the work is the area under the F–x line: W = ½k(x₂² − x₁²). Stretching a spring twice as far stores four times the energy.

Combined stiffness of several springs

1/k = 1/100 + 1/200 → k = 66.6667 N/m

Combined stiffness: 66.6667 N/m

Springs in series are softer than any one of them; springs in parallel are stiffer than any one of them.

Hooke's law is only linear up to the spring's elastic limit. Past that point the material yields, the extension stops being proportional to the force, and the spring does not return to its original length — so treat any answer far beyond a real spring's rated travel as an extrapolation, not a prediction.

What is the Hooke's Law Calculator?

Hooke's law says a spring pushes back in proportion to how far it is stretched: F = −kx, where k is the spring constant and x is the extension. The minus sign shows the force points back towards the rest position.

  • Solves F = kx for force, spring constant or extension
  • Elastic potential energy U = ½kx² shown in J, kJ, cal or kWh
  • Work done between two extensions using the area under the F-x line
  • Series and parallel spring combiner with add and remove rows
  • Spring rates in N/m, N/mm, N/cm, kN/m, lbf/in and lbf/ft
  • Runs completely in your browser with nothing sent anywhere

How to use the Hooke's Law Calculator

  1. 1

    Choose whether to solve for force, spring constant or extension.

  2. 2

    Enter the two known values and set their unit dropdowns, including the spring rate unit.

  3. 3

    Read the answer plus the stored elastic energy in the result tiles.

  4. 4

    Use the work panel to enter a start and end extension and see W = ½k(x₂² − x₁²).

  5. 5

    Add springs in the last panel and switch between series and parallel to get the combined stiffness.

About the Hooke's Law Calculator

The ByteTools Hooke's Law Calculator solves F = kx for the force, the spring constant or the extension, then reports the elastic potential energy stored in the spring from U = ½·k·x². Force can be entered in newtons, kilonewtons or pounds-force, stiffness in N/m, N/mm, N/cm, kN/m, lbf/in or lbf/ft, and extension in any of the usual length units.

Two extra panels go beyond the basic formula. One works out the work needed to stretch a spring between two extensions using W = ½k(x₂² − x₁²), which is the area under the force-extension line rather than a simple force times distance. The other combines any number of springs, adding stiffnesses in parallel and reciprocals in series so you can see how a stack behaves.

The whole calculation is JavaScript running in your browser. Nothing is uploaded, no account is needed and no history is kept, so answers appear as you type and the tool still works when you are offline, on a bench or in a lab without reliable wi-fi.

Frequently asked questions

What is Hooke's law?

Hooke's law states that the restoring force of a spring is proportional to its extension: F = −kx. The constant k is the stiffness in newtons per metre, and the minus sign means the force always acts back towards the spring's natural length.

How do you calculate the spring constant?

Divide the applied force by the resulting extension: k = F ÷ x. A spring that stretches 0.2 m under a 20 N load has a spring constant of 100 N/m, and a stiffer spring gives a larger number.

How much energy is stored in a stretched spring?

The elastic potential energy is U = ½·k·x², so a 100 N/m spring stretched 0.2 m stores 2 J. Because the extension is squared, stretching a spring twice as far stores four times as much energy.

Are springs in series stiffer or softer?

Springs joined end to end in series are softer than any one of them, following 1/k = Σ1/kᵢ. Springs side by side in parallel share the load and are stiffer than any single one, with k = Σkᵢ.

When does Hooke's law stop working?

It only holds up to the spring's elastic limit. Past that point the material yields, extension is no longer proportional to force, and the spring does not return to its original length, so any answer far beyond a real spring's rated travel is an extrapolation rather than a prediction.

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