Hooke's law and stiffness

What you'll learn

How force, extension and stiffness are linked for springs and wires.
How to use Hooke’s law: $F = k x$ .
How to find stiffness from a force-extension graph.
How elastic behaviour, plastic deformation and energy storage appear in this topic.

Starting point: forces and stretching

When you pull on a spring, wire or elastic material, you apply a force. Force is measured in newtons, N.

A force that stretches an object is called a tensile force. A force that squashes it is a compressive force. In this topic, we mainly talk about tensile forces because we are often measuring how much a spring or wire gets longer.

A load is the force applied to an object. In a school experiment, the load is usually the weight of hanging masses.

Weight is calculated using:

W = m g

where $W$ is weight in newtons, $m$ is mass in kilograms, and $g$ is gravitational field strength, usually taken as $kg−19.81\ \text{N kg}^{-1}$ .

Example

Converting hanging mass to force

A mass hanger and slotted masses have a total mass of 250 g. Find the force they apply.

Convert the mass into kilograms because $W = m g$ needs SI units: 250 g = 0.250 kg.
Substitute into the weight equation:
$\times 9.81$
Calculate the load:
$2.45\ \text{N}$

So the spring is being pulled down by about 2.45 N.

Common Mistake

Mass is not force

If you hang 200 g from a spring, the force is not 200 N. Convert mass to kilograms, then use $W = m g$ .

Length, original length and extension

The original length of an object is its length before the stretching force is applied. It is often written as $L_0$ .

The stretched length is the length while the force is acting. It is often written as $L$ .

Definition

Extension

The extension is the increase in length caused by the applied force:

x = L - L_0

Extension is measured in metres, m.

A typical practical setup uses a clamp stand, spring, ruler and hanging masses. A pointer helps you read the scale without parallax error.

A labelled setup for measuring spring extension using a clamp stand, spring, ruler, pointer and hanging masses

Tip

Use extension, not total length

Hooke’s law uses extension $x$ , not the total stretched length $L$ . Always subtract the original length first.

Hooke’s law

Many springs and wires behave very simply when the force is not too large: doubling the force doubles the extension, tripling the force triples the extension, and so on.

This is called direct proportionality. A directly proportional relationship gives a straight-line graph through the origin.

Key Idea

Hooke's law

Hooke’s law states that the force applied to an object is directly proportional to its extension, provided the limit of proportionality is not exceeded:

F = k x

Here:

$F$ is the applied force in newtons, N
$x$ is the extension in metres, m
$k$ is the spring constant, also called the stiffness, in newtons per metre, N m^-1

Definition

Stiffness

The stiffness of an object tells you how much force is needed to produce each metre of extension. A larger value of $k$ means a stiffer object.

A spring with a large stiffness does not stretch much for a given force. A spring with a small stiffness stretches more easily.

Example

Calculating extension

A spring has stiffness $150\ \text{N m}^{-1}$ . A force of 12.0 N is applied. Find the extension.

Start from Hooke’s law:
$F = k x$
Rearrange for extension:
$\frac{F}{k}$
Substitute the values:
$\frac{12.0}{150}$
Calculate the extension:
$0.0800\ \text{m}$

The extension is 0.0800 m, which is 8.00 cm.

Force-extension graphs

A force-extension graph shows how the applied force changes with extension.

If force is on the vertical axis and extension is on the horizontal axis, the gradient of the straight-line section is the stiffness:

\frac{\Delta F}{\Delta x}

The straight-line section is the region where Hooke’s law is obeyed.

A labelled force-extension graph showing the Hooke's law region, gradient as stiffness, and the limit of proportionality

The limit of proportionality is the point beyond which force is no longer directly proportional to extension. After this point, the graph is no longer a straight line through the origin.

Example

Finding stiffness from a graph

A force-extension graph has a straight-line section. Two points on the best-fit line are 2.0 cm, 3.2 N and 8.0 cm, 12.8 N. Find the stiffness.

Convert the extensions into metres:
$m2.0\ \text{cm} = 0.020\ \text{m}$ $m8.0\ \text{cm} = 0.080\ \text{m}$
Find the changes in force and extension:
$N\Delta F = 12.8 - 3.2 = 9.6\ \text{N}$ $m\Delta x = 0.080 - 0.020 = 0.060\ \text{m}$
Use gradient = stiffness:
$\frac{\Delta F}{\Delta x} = \frac{9.6}{0.060}$
Calculate the stiffness:
$160\ \text{N m}^{-1}$

Common Mistake

Wrong graph gradient

The gradient is $k$ only when force is on the y-axis and extension is on the x-axis. If the axes are swapped, the gradient is $1k\frac{1}{k}$ .

Elastic and plastic deformation

An object shows elastic deformation if it returns to its original shape and length when the force is removed.

An object shows plastic deformation if it has been permanently changed and does not return to its original length after unloading.

The elastic limit is the maximum force or extension an object can experience and still return to its original length afterwards.

Common Mistake

Limit of proportionality is not always the elastic limit

The limit of proportionality is where the graph stops being straight. The elastic limit is where permanent deformation begins. These are often close, but they are not exactly the same idea.

Example

Deciding whether deformation is elastic

A wire has original length 0.500 m. It is stretched to 0.535 m. When the force is removed, its length becomes 0.503 m. Decide whether the deformation was elastic or plastic.

Compare the final length after unloading with the original length.
The wire ends at 0.503 m, not 0.500 m, so it has not returned to its original length.
The permanent extension is:
$0.003\ \text{m}$

The deformation is plastic, with a permanent extension of 0.003 m.

Energy stored in a stretched spring

When you stretch a spring, you do work on it. That work is transferred into elastic strain energy, which is energy stored because the object has been deformed.

For any force-extension graph:

Key Idea

Energy from a graph

The work done stretching an object is the area under the force-extension graph.

For a Hooke’s law spring, the graph is a straight line through the origin, so the area is a triangle:

\frac{1}{2}Fx

Since $F = k x$ , you can also write:

\frac{1}{2}kx^2

Energy is measured in joules, J.

Example

Calculating elastic energy

A spring with stiffness $200\ \text{N m}^{-1}$ is stretched by 0.050 m. Find the elastic strain energy stored.

Use the Hooke’s law energy equation because the spring is assumed to be in the proportional region:
$\frac{1}{2}kx^2$
Substitute the values:
$\frac{1}{2} \times 200 \times (0.050)^2$
Calculate:
$0.250\ \text{J}$

So the spring stores 0.250 J of elastic strain energy.

Common Mistake

Do not use triangle area for every material

The equation $\frac{1}{2}Fx$ only applies when the force-extension graph is a straight line through the origin. For a curved graph, find the area under the curve instead.

Practical notes: measuring stiffness

To determine stiffness experimentally, you usually:

Measure the original length of the spring or wire.
Add known masses one at a time.
Convert each total mass into force using $W = m g$ .
Measure the new length each time.
Calculate extension using $x = L - L_0$ .
Plot force against extension.
Find the gradient of the straight-line section.

Good practical technique matters. Keep the ruler fixed, read at eye level, and use a fiducial marker or pointer to reduce parallax. Add masses gently so the spring does not oscillate too much.

For uncertainty, remember that extension comes from two length readings. If each ruler reading has an uncertainty of about ±1 mm, the uncertainty in extension is roughly ±2 mm.

Tip

Graph strategy

Use a large triangle on the best-fit straight line to calculate the gradient. Do not use a single plotted point unless the question specifically tells you to.

Stiffness is about the object, not just the material

A steel wire and a steel spring can have very different stiffness values, even though they may be made from the same material. Stiffness depends on the material and also on the object’s shape and dimensions.

In later materials work, you will meet Young modulus, which is a property of the material itself. For now, treat $k$ as the stiffness of a particular spring, wire or object.

Exam technique

In the exam

Check whether the question gives force or mass. If it gives mass, convert using $W = m g$ .
Check the graph axes before using the gradient. Force against extension gives gradient $k$ .
Use metres for extension, not cm or mm, when calculating stiffness in N m^-1.
Only apply $F = k x$ in the straight-line Hooke’s law region.
For energy, use the area under the force-extension graph; use $12Fx\frac{1}{2}Fx$ only for a straight line through the origin.

Self review

Check yourself

What is the difference between length and extension?
How would you find stiffness from a force-extension graph?
Why does Hooke’s law stop working after the limit of proportionality?

Hooke's law and stiffness

What you'll learn

Starting point: forces and stretching

Length, original length and extension

Hooke’s law

Force-extension graphs

Elastic and plastic deformation

Energy stored in a stretched spring

Practical notes: measuring stiffness

Stiffness is about the object, not just the material

Hooke's law and stiffness Revision Guide