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Physics

Spring Calculator

Use this spring calculator to solve Hooke's Law (F = k x) for any of its three variables: spring force, spring constant, or displacement. Switch to Energy mode to find the elastic potential energy stored in a compressed or stretched spring. Choose metric or imperial units, see the step-by-step working, and read an energy-vs-displacement chart that shows how stiffness shapes the energy curve.

Your details

Choose the quantity you want to calculate. The two remaining inputs become editable.
Stiffness of the spring. Higher values mean a stiffer spring.
N/m
Distance the spring is stretched or compressed from its natural (rest) length.
m
Spring forceMedium-stiffness spring
10

Force exerted by or on the spring (F = k x)

Spring constant200
Displacement0.05
Elastic potential energy0.25
Force unitN
Spring constant unitN/m
Displacement unitm
Energy unitJ
Force10
Potential energy0.25
01020000
Displacement (m)
Value
Displacement (m)Elastic PE (J)Force (N)
000
000.5
0.0101
0.010.011.5
0.010.012
0.010.022.5
0.020.023
0.020.033.5
0.020.044
0.020.054.5
0.030.065
0.030.085.5
0.030.096
0.030.116.5
0.040.127
0.040.147.5
0.040.168
0.040.188.5
0.050.29
0.050.239.5
0.050.2510
0.050.2810.5
0.060.311
0.060.3311.5
0.060.3612
0.060.3912.5
0.070.4213
0.070.4613.5
0.070.4914
0.070.5314.5
0.080.5615
0.080.615.5
0.080.6416
0.080.6816.5
0.090.7217
0.090.7717.5
0.090.8118
0.090.8618.5
0.10.919
0.10.9519.5
0.1120
  • Elastic PE (J)
  • Force (N)

A medium-stiffness spring (k = 200.00 N/m) stores 0.2500 J of elastic energy at this displacement.

  • A spring constant of 200.00 N/m means the spring requires 200.00 N of force per m of compression or extension.
  • At a displacement of 0.0500 m, the spring exerts a restoring force of 10.0000 N.
  • The elastic potential energy stored is 0.250000 J. Doubling the displacement would quadruple the stored energy, since PE grows with x squared.

Next stepTo find the oscillation period of a mass on this spring, use T = 2pi sqrt(m/k) in our Simple Harmonic Motion calculator.

Formula

F=kx(Hooke’s Law),PE=12kx2(Elastic potential energy)F = k\,x \quad (\text{Hooke's Law}), \qquad PE = \tfrac{1}{2}\,k\,x^{2} \quad (\text{Elastic potential energy})

Worked example

A spring with k = 200 N/m is compressed by x = 0.05 m. Force: F = 200 x 0.05 = 10 N. Elastic PE: PE = 0.5 x 200 x 0.05^2 = 0.25 J. The restoring force acts in the opposite direction to the compression: -10 N.

What is Hooke's Law?

Hooke's Law describes the behavior of an elastic spring: the restoring force it produces is directly proportional to how far it has been stretched or compressed from its natural length. In equation form: F = k x, where F is the force in newtons, k is the spring constant in N/m (a measure of stiffness), and x is the displacement in metres. The law is named after the English physicist Robert Hooke, who stated it in 1678. It holds accurately for small to moderate displacements - once a spring is stretched past its elastic limit it deforms permanently and the linear relationship breaks down.

How to use this spring calculator

Select what you want to solve for from the "Solve for" menu: spring force (F), spring constant (k), displacement (x), or elastic potential energy (PE). Enter the two known values and the result appears instantly. Switch between metric (newtons and metres) and imperial (pound-force and inches) at any time. The "Show your work" panel walks through the arithmetic step by step using your actual numbers, and the chart below plots both the force-displacement line and the parabolic energy curve up to twice your entered displacement so you can see how the values change beyond the current point.

Spring constant, displacement, and elastic potential energy explained

The spring constant k is the defining characteristic of a spring. A spring with k = 1,000 N/m needs 1,000 N (about 100 kg of weight at Earth's surface) to compress or extend it by 1 metre - or just 1 N per millimetre. Displacement x is measured from the natural (un-loaded) length, and it does not matter whether the spring is compressed or extended: the magnitude of the restoring force is the same. Elastic potential energy is the energy stored by the deformation: PE = (1/2) k x squared. Because of the x squared term, doubling the displacement stores four times as much energy, which is why compressed springs can release energy very rapidly and why care is needed with heavily loaded springs.

Sign conventions and the restoring force

In the standard physics convention the restoring force has a negative sign relative to displacement: F_restoring = -k x. This means the spring always pushes or pulls back toward its rest position. Engineering practice often drops the sign and works with the magnitude F = k x, interpreting direction from context. This calculator uses the engineering magnitude convention for the displayed result. For simple harmonic motion, the signed form matters: the acceleration of a mass m on a spring is a = -k x / m, giving an angular frequency of omega = sqrt(k/m) and a period T = 2 pi sqrt(m/k).

Typical spring constant (k) values by application

ApplicationTypical k (N/m)Category
Mattress coil spring5 - 50 Very soft
Pen click spring50 - 300 Soft
Suspension spring (bicycle)5,000 - 20,000 Medium
Suspension spring (car)15,000 - 80,000 Stiff
Valve spring (engine)20,000 - 100,000 Stiff
Industrial press spring100,000 - 1,000,000 Very stiff

Approximate stiffness values for common spring types. Actual values vary widely by design and material.

Frequently asked questions

What is the spring constant (k)?

The spring constant k measures a spring's stiffness. It tells you how many newtons of force the spring produces (or requires) for each metre of stretch or compression. A larger k means a stiffer spring. Typical values range from a few N/m for soft mattress springs to hundreds of thousands of N/m for industrial press springs.

What is the formula for spring force?

The spring force is given by Hooke's Law: F = k x, where k is the spring constant in N/m and x is the displacement in metres. For example, a spring with k = 500 N/m compressed by 0.1 m exerts a restoring force of 50 N. In imperial units the same formula applies with k in lbf/in and x in inches.

How is elastic potential energy calculated?

Elastic potential energy (PE) stored in a spring is PE = (1/2) k x squared. It grows with the square of the displacement, so doubling the stretch stores four times as much energy. At x = 0.05 m with k = 200 N/m, PE = 0.5 x 200 x 0.0025 = 0.25 J.

What is Hooke's Law and when does it apply?

Hooke's Law states that the force needed to extend or compress a spring is proportional to the displacement, within the spring's elastic limit. It applies accurately for small deformations. Beyond the elastic limit the spring deforms permanently and the law no longer holds. Most practical spring calculations use Hooke's Law because springs are designed to operate within their elastic range.

How do I convert between metric and imperial spring units?

1 lbf/in = 175.127 N/m, and 1 inch = 0.0254 m, so a spring rated at 10 lbf/in is equivalent to about 1,751 N/m. For force, 1 lbf = 4.44822 N. This calculator handles the unit arithmetic automatically when you switch the unit system selector.

Sources

Written by Dr. Tomás Okafor, PhD Physicist · Lagos, Nigeria

Physicist specializing in classical mechanics, bringing 17 years of research and applied dynamics expertise to every calculator he reviews.

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