Hooke's law

Extension and compression

Extension happens when an object increases in length, and compression happens when it decreases in length. The extension of an elastic object when a force is applied, such as a spring, is described by Hooke's law:

force = spring constant × extension

F = k \: x

This is when:

  • force (F) is measured in newtons (N)
  • spring constant (k) is measured in newtons per metre (N/m)
  • extension, or increase in length (x), is measured in metres (m)

Example

A force of 3 N is applied to a spring. The spring stretches reversibly by 0.15 m - the fact that the string stretches reversibly means that it will go back to its normal shape after the force has been removed. Calculate the spring constant.

First rearrange F = k \: x to find k:

k = \frac{F}{x}

Then calculate using the values in the question:

k = 3 \div 0.15

k = 20 \: N/m

Limit of proportionality

Spring constant is a measure of the stiffness and strength of a spring. The limit of proportionality refers to the point beyond which Hooke's law is no longer true when stretching a material. The elastic limit of a material is the furthest amount it can be stretched or deformed without being able to return to its previous shape. Once a material has gone past its elastic limit, its deformation is said to be inelastic.

The higher the spring constant, the stiffer the spring. The spring constant is not the same value for different elastic objects. For a given spring and other elastic objects, the extension is directly proportional to the force applied. For example, if the force is doubled, the extension doubles. This works until the limit of proportionality is exceeded.

When an elastic object is stretched beyond its limit of proportionality the object does not return to its original length when the force is removed. In this instance, the relationship between force and extension changes from being linear, or directly proportional, to being non-linear.

Force-extension graphs

A force extension graph. Linear section drawn from origin to occupy half of graph area. Non linear section has decreasing gradient. Change from linear to non-linear is marked and labelled.

Linear extension and elastic deformation can be seen below the limit of proportionality.

Non-linear extension and inelastic deformation can be seen above the limit of proportionality. The limit of proportionality is also described as the elastic limit. The gradient of a force-extension graph before the limit of proportionality is equal to the spring constant.

Energy stored in a spring

Work is done when a spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.

The elastic potential energy stored can be calculated using the equation:

E = \frac{1}{2}kx^{2}

This is when:

  • elastic potential energy (E) is measured in joules (J)
  • spring constant (k) is measured in newtons per metre (N/m)
  • extension, referring to the increase in length (x), is measured in metres (m)
curriculum-key-fact
This equation also works for the reduction in length when a spring is compressed.

Example

A spring has a spring constant (k) of 3 N/m. It is stretched until it is extended by 50 cm. Calculate the elastic potential energy stored by the spring, assuming it is not stretched beyond the limit of proportionality.

First convert centimetres to metres:

50 cm = 50 ÷ 100 = 0.5 m

Then calculate using the values in the question:

E = \frac{1}{2}kx^{2}

E = \frac{1}{2} \times 3 \times 0.5^{2}

E = 1.5 \times 0.25

E = 0.375 \:J

Question

A spring is compressed by 0.15 m. It has a spring constant of 80 N/m. Calculate the elastic potential energy stored by the spring.

E = \frac{1}{2}kx^{2}

E = \frac{1}{2} \times 80 \times 0.15^{2}

E = 40 \times 0.0225

E = 0.90 \:J