Hooke's Law: Spring Force and Elastic Energy

Calculates the force of a spring as F = spring constant × extension: the force grows in proportion to the stretch. The stored elastic energy is ½ × spring constant × extension².

Explanation

Stretch or squash a spring and it pulls back towards its natural length. Hooke's law says that this restoring force grows in proportion to how far the spring has been stretched.

F = k x

$F$ is the spring force, in newtons (N)
$k$ is the spring constant, in newtons per metre (N/m)
$x$ is the extension from the natural length, in metres (m)

The spring constant measures stiffness: $k = 200$ N/m means it takes 200 N to stretch the spring by one metre.

As you pull, the force needed climbs steadily from zero to $k x$ , so on average it is $\frac{1}{2} k x$ . Multiply that average by the distance $x$ and you get the elastic energy the spring has stored.

E = \dfrac{1}{2} k x^2

Example

The defaults are a spring constant of 200 N/m and an extension of 0.1 m.

F = 200 \times 0.1 = 20\,\mathrm{N}

E = \dfrac{1}{2} \times 200 \times 0.1^2 = 1\,\mathrm{J}

The spring pulls with 20 N and holds 1 J of elastic energy. Let go and that joule becomes the kinetic energy of whatever is attached.

Notes

Extensions go in metres. A stretch measured as 10 cm is 0.1 m. Leave it in centimetres and the force comes out 100 times too big.

The spring pulls back against the stretch, so written with its direction the law reads $F = -k x$ . This calculator gives the size of the force.

Hooke's law only holds within the elastic limit. Pull a spring too far and it stays deformed, and the formula stops describing it.

The spring constant must be greater than zero. Zero or a negative value gives an error.