Force-extension graphs and elastic energy

What you'll learn

How to interpret a force-extension graph for a spring, wire or rubber band.
How to use Hooke’s law, $F = k x$ , and find the spring constant from a graph.
How to calculate elastic strain energy from the area under a graph.
How to recognise elastic deformation, plastic deformation and energy losses.

Starting point: forces, length and extension

When you stretch an object, such as a spring or a metal wire, its length increases. In this topic we usually measure:

the original length before loading
the new length after a force is applied
the extension, which is the increase in length

For a hanging mass, the stretching force is usually its weight:

W = m g

where $m$ is the mass in kg and $g$ is the gravitational field strength, about 9.81 N kg^-1 near Earth’s surface.

Definition

Extension

The extension $x$ of an object is the change in its length:

x=L-L_0

where $L_0$ is the original length and $L$ is the stretched length. Extension is measured in metres, m.

Common Mistake

Using length instead of extension

A force-extension graph uses extension on the horizontal axis, not total length. If a spring grows from 12.0 cm to 15.0 cm, the extension is 3.0 cm, not 15.0 cm.

Example

Finding the extension and load

A spring is originally 18.0 cm long. A 250 g mass is hung from it, and its new length is 21.5 cm. Find the extension and the stretching force.

Convert the lengths into metres and subtract the original length from the stretched length: $mx=0.215\ \text{m}-0.180\ \text{m}=0.035\ \text{m}$ .
Convert the mass into kg so it can be used in $W = m g$ : $kgm=250\ \text{g}=0.250\ \text{kg}$ .
Calculate the load: $NW=mg=0.250\ \text{kg}\times 9.81\ \text{N kg}^{-1}=2.45\ \text{N}$ .
This measurement would be plotted as the point $mx=0.035\ \text{m}$ , $NF=2.45\ \text{N}$ on a force-extension graph.

Hooke’s law and the spring constant

For many springs and wires, small extensions are very predictable: doubling the force doubles the extension. That is Hooke’s law.

Definition

Hooke’s law

Hooke’s law states that the force applied to a spring or wire is directly proportional to its extension, provided the limit of proportionality has not been exceeded:

F = k x

Here $F$ is force in newtons, N, $x$ is extension in metres, m, and $k$ is the spring constant in N m^-1.

The spring constant $k$ tells you how stiff the object is. A larger $k$ means more force is needed for the same extension.

On a force-extension graph, the straight-line section has gradient:

k=ΔFΔxk=\frac{\Delta F}{\Delta x}

Key Idea

Gradient means stiffness

For a force-extension graph, the gradient of the straight-line section is the spring constant $k$ . A steeper line means a stiffer spring or wire.

Example

Finding the spring constant from a graph

A spring has a straight-line force-extension graph. At an extension of 0.040 m, the force is 8.0 N. Find the spring constant and predict the extension when the force is 5.0 N.

Use the gradient of the straight-line graph: $k=ΔFΔxk=\frac{\Delta F}{\Delta x}$ .
Substitute the graph values: $m−1k=\frac{8.0\ \text{N}}{0.040\ \text{m}}=200\ \text{N m}^{-1}$ .
Rearrange Hooke’s law to find extension: $x=Fkx=\frac{F}{k}$ .
Substitute the new force: $mx=\frac{5.0\ \text{N}}{200\ \text{N m}^{-1}}=0.025\ \text{m}$ .

Tip

Choosing a gradient triangle

When finding $k$ from a graph, use a large triangle drawn on the best-fit straight line. Do not just divide one plotted point by another if the data are noisy.

Reading a force-extension graph

A typical force-extension graph has force on the vertical axis and extension on the horizontal axis. For a spring obeying Hooke’s law, the graph is a straight line through the origin. Once the material stops behaving proportionally, the graph curves.

Force-extension graph showing Hooke's law region, limit of proportionality and elastic strain energy area

Definition

Limits and deformation

The limit of proportionality is the point beyond which force is no longer directly proportional to extension.
The elastic limit is the point beyond which the object will not return to its original length when the force is removed.
Elastic deformation means the object returns to its original shape and length after the load is removed.
Plastic deformation means the object has a permanent change in shape or length after the load is removed.

The limit of proportionality and elastic limit are often close together, but they are not the same idea. A material can stop being proportional before it becomes permanently deformed.

Common Mistake

Do not overuse Hooke’s law

Only use $F = k x$ while the graph is straight and passes through the origin. Beyond the limit of proportionality, $k$ is no longer constant for that object.

Work done and elastic strain energy

When you stretch a spring, you do work on it. That energy is transferred into the spring as elastic strain energy, as long as the deformation is elastic.

Definition

Elastic strain energy

Elastic strain energy is the energy stored in a stretched or compressed object due to its deformation. It is measured in joules, J.

For any force-extension graph:

Key Idea

Area under the graph

The work done stretching a material is equal to the area under its force-extension graph.

This is because the force may not be constant while the object stretches. For a small extra extension, the work done is approximately force times extra extension. Adding all these small strips gives the area under the graph.

For a spring obeying Hooke’s law, the graph is a triangle, so:

Ee=12FxE_e=\frac{1}{2}Fx

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

Ee=12kx2E_e=\frac{1}{2}kx^2

or:

Ee=F22kE_e=\frac{F^2}{2k}

These are all equivalent for the straight-line Hooke’s law region.

Common Mistake

Forgetting the factor of one half

For a Hooke’s law spring, the force increases from zero to $F$ , so the average force is $F2\frac{F}{2}$ . That is why elastic strain energy is $12Fx\frac{1}{2}Fx$ , not $F x$ .

Example

Calculating elastic strain energy

A spring has spring constant 250 N m^-1 and is stretched by 6.0 cm. Calculate the elastic strain energy stored.

Convert the extension into metres: $mx=6.0\ \text{cm}=0.060\ \text{m}$ .
Choose the equation using $k$ and $x$ : $Ee=12kx2E_e=\frac{1}{2}kx^2$ .
Substitute carefully: $m)2E_e=\frac{1}{2}\times 250\ \text{N m}^{-1}\times \left(0.060\ \text{m}\right)^2$ .
Calculate the energy: $JE_e=0.45\ \text{J}$ .

Curved force-extension graphs

Not all materials have a straight force-extension graph. Rubber bands, biological tissues and materials stretched beyond their proportional region often give curved graphs.

For curved graphs, you cannot use $12Fx\frac{1}{2}Fx$ unless the graph is actually a straight line from the origin. Instead, estimate the area under the graph by counting squares or by splitting the area into trapezia.

Example

Estimating work done from graph data

A material is stretched and the following values are read from a curved force-extension graph: at 0 m, force is 0 N; at 0.020 m, force is 4.0 N; at 0.040 m, force is 9.0 N; at 0.060 m, force is 15.0 N. Estimate the work done from 0 m to 0.060 m.

Split the area under the graph into three trapezia, each with width 0.020 m.
Calculate the first trapezium: $J\frac{1}{2}\left(0+4.0\right)\text{N}\times 0.020\ \text{m}=0.040\ \text{J}$ .
Calculate the second trapezium: $J\frac{1}{2}\left(4.0+9.0\right)\text{N}\times 0.020\ \text{m}=0.130\ \text{J}$ .
Calculate the third trapezium: $J\frac{1}{2}\left(9.0+15.0\right)\text{N}\times 0.020\ \text{m}=0.240\ \text{J}$ .
Add the areas: $JW=0.040\ \text{J}+0.130\ \text{J}+0.240\ \text{J}=0.410\ \text{J}$ , so the work done is about 0.41 J.

Loading, unloading and hysteresis

Loading means increasing the force on the material. Unloading means removing the force. For an ideal spring within its elastic limit, the unloading graph follows the same line back to the origin, so the stored energy is returned.

Rubber bands often behave differently: the unloading curve is below the loading curve. This means less energy is returned than was put in.

Force-extension graph for a rubber band showing loading, unloading and energy dissipated

Definition

Hysteresis

Hysteresis is when the loading and unloading curves are different. The area between the curves is the energy dissipated, usually transferred to thermal energy in the material and surroundings.

Example

Finding energy dissipated in a loading cycle

A rubber band is stretched and released. The area under the loading curve is 1.20 J. The area under the unloading curve is 0.75 J. Find the energy dissipated.

Interpret the loading area as the work done on the rubber band: $JE_{\text{in}}=1.20\ \text{J}$ .
Interpret the unloading area as the energy returned by the rubber band: $JE_{\text{returned}}=0.75\ \text{J}$ .
Subtract the returned energy from the input energy: $JE_{\text{dissipated}}=1.20\ \text{J}-0.75\ \text{J}=0.45\ \text{J}$ .

Practical graph skills

In a typical lab experiment, you might hang masses from a spring or wire and measure the extension for each load. To get reliable force-extension data:

measure the original length before adding masses
convert mass to force using $W = m g$
wait for the object to stop oscillating before reading the ruler
use a fiducial marker or set square to reduce parallax error
plot force on the vertical axis and extension on the horizontal axis
draw a best-fit line or curve, not dot-to-dot lines

If each length reading has an uncertainty of about 1 mm, the extension may have a larger uncertainty because it is found by subtracting two readings.

Tip

Graph marks are often easy marks

Label axes with quantities and units, use SI units, plot points accurately, and take the gradient from a large triangle on the best-fit straight section. If the graph curves, describe the trend rather than forcing a straight line.

Energy transfers using springs

Elastic strain energy can be transferred into other forms. For example, a stretched spring might launch a trolley, so elastic strain energy becomes kinetic energy. If there are no significant losses, you can equate the energies:

12kx2=12mv2\frac{1}{2}kx^2=\frac{1}{2}mv^2

In real situations, some energy is often dissipated due to friction, air resistance or internal heating.

Exam technique

In the exam

Check whether the graph is straight through the origin before using $F = k x$ or $Ee=12FxE_e=\frac{1}{2}Fx$ .
Convert extensions into metres and masses into kg before substituting into equations.
For curved graphs, use the area under the graph: count squares or split the area into trapezia.
State units clearly: spring constant in N m^-1 and energy in J.

Self review

Check yourself

What does the gradient of a straight force-extension graph tell you?
Why is the elastic strain energy in a Hooke’s law spring $12Fx\frac{1}{2}Fx$ rather than $F x$ ?
What does the area between loading and unloading curves represent?

Force-extension graphs and elastic energy

What you'll learn

Starting point: forces, length and extension

Hooke’s law and the spring constant

Reading a force-extension graph

Work done and elastic strain energy

Curved force-extension graphs

Loading, unloading and hysteresis

Practical graph skills

Energy transfers using springs

Force-extension graphs and elastic energy Revision Guide