Definition

Work done

The energy transferred when a force moves an object through a distance in the direction of the force.

Definition

Elastic potential energy

The energy held in the elastic store of an object that has been stretched, compressed or bent elastically.

To stretch a spring, a force must move the end of the spring through a distance, so work is done on the spring.
The work done in stretching a spring is equal to the energy transferred to it.
When the spring is stretched elastically, this energy is held in the spring's elastic store.
When the spring is released, energy in its elastic store is transferred to other stores, such as the kinetic store of a ball fired from a toy launcher.
1. A stretched catapult or a drawn bow transfers energy from its elastic store to the kinetic store of the stone or the arrow when it is released.
If a spring is stretched beyond its elastic limit, part of the work done permanently rearranges the material and warms it, so not all of the energy can be returned.

Why the force changes during a stretch

The force needed to stretch a spring increases as the extension increases.
At the start of the stretch the force is zero, and at an extension $x$ it has risen to $k x$ .
For a spring obeying Hooke's law the force rises steadily, so the average force during the stretch is half of the final force.
The work done is therefore not the final force multiplied by the extension, but half of that value.

The energy equation

The energy transferred in stretching a spring is given by an equation that you do not have to memorise, but you must be able to choose it and use it.
In words, $constant×(extension)2\text{energy transferred in stretching}=\frac{1}{2}\times\text{spring constant}\times(\text{extension})^2$ .
In symbols, the equation is $E=12kx2E=\frac{1}{2}kx^2$ $E = \frac{1}{2} k x^{2}$ .
1. $E$ is the energy transferred, equal to the work done, in joules, $J\text{J}$ .
2. $k$ is the spring constant in newtons per metre, $N/m\text{N/m}$ .
3. $x$ is the extension in metres, $m\text{m}$ .
The extension is squared, so doubling the extension makes the energy stored four times as large.
The equation applies only up to the limit of proportionality, because it assumes that the force rises in proportion to the extension.
Rearranging the equation gives $k=2Ex2k=\dfrac{2E}{x^2}$ for the spring constant and $x=2Ekx=\sqrt{\dfrac{2E}{k}}$ for the extension.

Work done from a force-extension graph

On a graph of force against extension, the area under the line equals the work done in stretching.
This works because work done is force multiplied by distance, and each thin strip under the graph is a small force multiplied by a small extension.
For the straight section of a spring's graph, the area is a triangle, so $done=12×force×extension\text{work done}=\frac{1}{2}\times\text{force}\times\text{extension}$ .
Substituting $F = k x$ into the triangle area gives $12×kx×x=12kx2\frac{1}{2}\times kx\times x=\frac{1}{2}kx^2$ , which is the energy equation.
For a curved graph, such as the one for a rubber band, no simple equation applies, so the area is found by counting squares under the curve.
1. Count each square that is at least half under the curve as a whole square, and ignore squares that are less than half under it.
2. Find the energy represented by one square by multiplying the force for one square by the extension for one square.
3. Multiply the number of squares by the energy for one square to give the total work done.
The unit of area on this graph is the newton metre, $m\text{N}\,\text{m}$ , and $m1\ \text{N}\,\text{m}$ is equal to $J1\ \text{J}$ .

Example

Energy stored in a stretched spring

A spring with a spring constant of $N/m40\ \text{N/m}$ is stretched by $cm15\ \text{cm}$ .

x=15\ \text{cm}=0.15\ \text{m}

E=\frac{1}{2}kx^2=\frac{1}{2}\times40\ \text{N/m}\times(0.15\ \text{m})^2

E=\frac{1}{2}\times40\times0.0225=0.45\ \text{J}

The work done in stretching the spring is $J0.45\ \text{J}$ , and this energy is held in its elastic store.

Example

Energy from the area under a straight line

A force-extension graph for a spring is a straight line from the origin to the point where the force is $N6.0\ \text{N}$ and the extension is $m0.12\ \text{m}$ .

\text{work done}=\frac{1}{2}\times\text{force}\times\text{extension}=\frac{1}{2}\times6.0\ \text{N}\times0.12\ \text{m}=0.36\ \text{J}

The area of the triangle under the line shows that $J0.36\ \text{J}$ of work is done.

Example

Extension from the energy stored

A spring with a spring constant of $N/m800\ \text{N/m}$ stores $J4.0\ \text{J}$ in its elastic store.

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2\times4.0\ \text{J}}{800\ \text{N/m}}}=\sqrt{0.010}=0.10\ \text{m}

The spring is stretched by $m0.10\ \text{m}$ , which is $cm10\ \text{cm}$ .

Example

Work done by counting squares

A curved force-extension graph for a rubber band has squares that each represent $N0.5\ \text{N}$ of force and $m0.01\ \text{m}$ of extension.

\text{energy for one square}=0.5\ \text{N}\times0.01\ \text{m}=0.005\ \text{J}

There are $48$ squares under the curve, counting squares that are at least half covered.

\text{work done}=48\times0.005\ \text{J}=0.24\ \text{J}

The work done in stretching the rubber band is about $J0.24\ \text{J}$ .

Practical

Work done in stretching a rubber band

Aim: to plot a force-extension graph for a rubber band stretched by hanging weights and to find the work done in stretching it from the area under the graph.
Apparatus: a clamp stand with two bosses and clamps, a G-clamp, a thick rubber band, a $g100\ \text{g}$ mass hanger with $g100\ \text{g}$ slotted masses, a metre rule, a set square, a strip of tape as a pointer, graph paper, a tray of sand and eye protection.
Method:
- Fix the clamp stand to the bench with the G-clamp, loop the rubber band over the upper clamp and hold the metre rule vertically beside it in the lower clamp.
- Fold the tape around the bottom of the band so that it sticks out across the scale, and record its position with no load, reading at eye level against the set square.
- Hang the $g100\ \text{g}$ mass hanger on the band, which applies a force of $N1\ \text{N}$ , wait $s30\ \text{s}$ and record the pointer position.
- Add $g100\ \text{g}$ masses one at a time up to a total force of $N8\ \text{N}$ , waiting $s30\ \text{s}$ each time before recording the position.
- Subtract the unloaded reading from each loaded reading and convert every extension into metres.
- Repeat the loading with a fresh band of the same type twice more and calculate a mean extension for each force.
- Plot force on the vertical axis against extension on the horizontal axis, choosing scales that fill at least half of the graph paper, and draw a smooth curve of best fit through the origin.
- Count the squares under the curve up to the largest extension, find the energy for one square, and multiply to find the total work done.
Results: the points form a curve rather than a straight line, and a typical thick rubber band loaded to $N8\ \text{N}$ needs a few tenths of a joule of work to stretch it.
Conclusion: the work done in stretching the band equals the area under its force-extension graph, and counting squares is needed because the graph is curved, so $E=12kx2E=\frac{1}{2}kx^2$ cannot be used.
Watch out: plot force on the vertical axis so that the area gives the work done, convert extensions from centimetres to metres before working out the energy for one square, and keep the waiting time the same because rubber keeps slowly stretching under a steady load.
Safety: wear eye protection because a rubber band can snap and whip back, place the tray of sand beneath the masses, and keep your feet clear of the hanging load.

Common Mistake

Do not calculate the work done as the final force multiplied by the extension, because the force rises from zero during the stretch.
Do not forget to square the extension in $E=12kx2E=\frac{1}{2}kx^2$ , and convert it into metres before squaring it.
Do not use $E=12kx2E=\frac{1}{2}kx^2$ for a rubber band or for a spring beyond its limit of proportionality, because the force is no longer proportional to the extension.

Exam technique

Work done in stretching

Choose the equation $E=12kx2E=\frac{1}{2}kx^2$ , write it down and then substitute, with the extension in metres.
Show the squared term as a separate value, such as $0.15^2=0.0225$ , so that every stage of your working can be credited.
When a graph is given, state that the work done is the area under the force-extension graph and show how you found the area.
Give energy answers in joules, and name the elastic store when asked where the energy goes.

Self review

State which energy store gains energy when a spring is stretched elastically.
Write the equation for the energy transferred in stretching a spring.
Calculate the energy stored in a spring with a spring constant of $N/m60\ \text{N/m}$ that is stretched by $m0.20\ \text{m}$ .
Explain what the area under a force-extension graph represents.
Describe how to find the work done from a curved force-extension graph.

Definition

Work done

The energy transferred when a force moves an object through a distance in the direction of the force.

Definition

Elastic potential energy

The energy held in the elastic store of an object that has been stretched, compressed or bent elastically.

To stretch a spring, a force must move the end of the spring through a distance, so work is done on the spring.
The work done in stretching a spring is equal to the energy transferred to it.
When the spring is stretched elastically, this energy is held in the spring's elastic store.
When the spring is released, energy in its elastic store is transferred to other stores, such as the kinetic store of a ball fired from a toy launcher.
1. A stretched catapult or a drawn bow transfers energy from its elastic store to the kinetic store of the stone or the arrow when it is released.
If a spring is stretched beyond its elastic limit, part of the work done permanently rearranges the material and warms it, so not all of the energy can be returned.

Why the force changes during a stretch

The force needed to stretch a spring increases as the extension increases.
At the start of the stretch the force is zero, and at an extension $x$ it has risen to $k x$ .
For a spring obeying Hooke's law the force rises steadily, so the average force during the stretch is half of the final force.
The work done is therefore not the final force multiplied by the extension, but half of that value.

The energy equation

The energy transferred in stretching a spring is given by an equation that you do not have to memorise, but you must be able to choose it and use it.
In words, $constant×(extension)2\text{energy transferred in stretching}=\frac{1}{2}\times\text{spring constant}\times(\text{extension})^2$ .
In symbols, the equation is $E=12kx2E=\frac{1}{2}kx^2$ $E = \frac{1}{2} k x^{2}$ .
1. $E$ is the energy transferred, equal to the work done, in joules, $J\text{J}$ .
2. $k$ is the spring constant in newtons per metre, $N/m\text{N/m}$ .
3. $x$ is the extension in metres, $m\text{m}$ .
The extension is squared, so doubling the extension makes the energy stored four times as large.
The equation applies only up to the limit of proportionality, because it assumes that the force rises in proportion to the extension.
Rearranging the equation gives $k=2Ex2k=\dfrac{2E}{x^2}$ for the spring constant and $x=2Ekx=\sqrt{\dfrac{2E}{k}}$ for the extension.

Work done from a force-extension graph

On a graph of force against extension, the area under the line equals the work done in stretching.
This works because work done is force multiplied by distance, and each thin strip under the graph is a small force multiplied by a small extension.
For the straight section of a spring's graph, the area is a triangle, so $done=12×force×extension\text{work done}=\frac{1}{2}\times\text{force}\times\text{extension}$ .
Substituting $F = k x$ into the triangle area gives $12×kx×x=12kx2\frac{1}{2}\times kx\times x=\frac{1}{2}kx^2$ , which is the energy equation.
For a curved graph, such as the one for a rubber band, no simple equation applies, so the area is found by counting squares under the curve.
1. Count each square that is at least half under the curve as a whole square, and ignore squares that are less than half under it.
2. Find the energy represented by one square by multiplying the force for one square by the extension for one square.
3. Multiply the number of squares by the energy for one square to give the total work done.
The unit of area on this graph is the newton metre, $m\text{N}\,\text{m}$ , and $m1\ \text{N}\,\text{m}$ is equal to $J1\ \text{J}$ .

Example

Energy stored in a stretched spring

A spring with a spring constant of $N/m40\ \text{N/m}$ is stretched by $cm15\ \text{cm}$ .

x=15\ \text{cm}=0.15\ \text{m}

E=\frac{1}{2}kx^2=\frac{1}{2}\times40\ \text{N/m}\times(0.15\ \text{m})^2

E=\frac{1}{2}\times40\times0.0225=0.45\ \text{J}

The work done in stretching the spring is $J0.45\ \text{J}$ , and this energy is held in its elastic store.

Example

Energy from the area under a straight line

A force-extension graph for a spring is a straight line from the origin to the point where the force is $N6.0\ \text{N}$ and the extension is $m0.12\ \text{m}$ .

\text{work done}=\frac{1}{2}\times\text{force}\times\text{extension}=\frac{1}{2}\times6.0\ \text{N}\times0.12\ \text{m}=0.36\ \text{J}

The area of the triangle under the line shows that $J0.36\ \text{J}$ of work is done.

Example

Extension from the energy stored

A spring with a spring constant of $N/m800\ \text{N/m}$ stores $J4.0\ \text{J}$ in its elastic store.

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2\times4.0\ \text{J}}{800\ \text{N/m}}}=\sqrt{0.010}=0.10\ \text{m}

The spring is stretched by $m0.10\ \text{m}$ , which is $cm10\ \text{cm}$ .

Example

Work done by counting squares

A curved force-extension graph for a rubber band has squares that each represent $N0.5\ \text{N}$ of force and $m0.01\ \text{m}$ of extension.

\text{energy for one square}=0.5\ \text{N}\times0.01\ \text{m}=0.005\ \text{J}

There are $48$ squares under the curve, counting squares that are at least half covered.

\text{work done}=48\times0.005\ \text{J}=0.24\ \text{J}

The work done in stretching the rubber band is about $J0.24\ \text{J}$ .

Practical

Work done in stretching a rubber band

Aim: to plot a force-extension graph for a rubber band stretched by hanging weights and to find the work done in stretching it from the area under the graph.
Apparatus: a clamp stand with two bosses and clamps, a G-clamp, a thick rubber band, a $g100\ \text{g}$ mass hanger with $g100\ \text{g}$ slotted masses, a metre rule, a set square, a strip of tape as a pointer, graph paper, a tray of sand and eye protection.
Method:
- Fix the clamp stand to the bench with the G-clamp, loop the rubber band over the upper clamp and hold the metre rule vertically beside it in the lower clamp.
- Fold the tape around the bottom of the band so that it sticks out across the scale, and record its position with no load, reading at eye level against the set square.
- Hang the $g100\ \text{g}$ mass hanger on the band, which applies a force of $N1\ \text{N}$ , wait $s30\ \text{s}$ and record the pointer position.
- Add $g100\ \text{g}$ masses one at a time up to a total force of $N8\ \text{N}$ , waiting $s30\ \text{s}$ each time before recording the position.
- Subtract the unloaded reading from each loaded reading and convert every extension into metres.
- Repeat the loading with a fresh band of the same type twice more and calculate a mean extension for each force.
- Plot force on the vertical axis against extension on the horizontal axis, choosing scales that fill at least half of the graph paper, and draw a smooth curve of best fit through the origin.
- Count the squares under the curve up to the largest extension, find the energy for one square, and multiply to find the total work done.
Results: the points form a curve rather than a straight line, and a typical thick rubber band loaded to $N8\ \text{N}$ needs a few tenths of a joule of work to stretch it.
Conclusion: the work done in stretching the band equals the area under its force-extension graph, and counting squares is needed because the graph is curved, so $E=12kx2E=\frac{1}{2}kx^2$ cannot be used.
Watch out: plot force on the vertical axis so that the area gives the work done, convert extensions from centimetres to metres before working out the energy for one square, and keep the waiting time the same because rubber keeps slowly stretching under a steady load.
Safety: wear eye protection because a rubber band can snap and whip back, place the tray of sand beneath the masses, and keep your feet clear of the hanging load.

Common Mistake

Do not calculate the work done as the final force multiplied by the extension, because the force rises from zero during the stretch.
Do not forget to square the extension in $E=12kx2E=\frac{1}{2}kx^2$ , and convert it into metres before squaring it.
Do not use $E=12kx2E=\frac{1}{2}kx^2$ for a rubber band or for a spring beyond its limit of proportionality, because the force is no longer proportional to the extension.

Exam technique

Work done in stretching

Choose the equation $E=12kx2E=\frac{1}{2}kx^2$ , write it down and then substitute, with the extension in metres.
Show the squared term as a separate value, such as $0.15^2=0.0225$ , so that every stage of your working can be credited.
When a graph is given, state that the work done is the area under the force-extension graph and show how you found the area.
Give energy answers in joules, and name the elastic store when asked where the energy goes.

Self review

State which energy store gains energy when a spring is stretched elastically.
Write the equation for the energy transferred in stretching a spring.
Calculate the energy stored in a spring with a spring constant of $N/m60\ \text{N/m}$ that is stretched by $m0.20\ \text{m}$ .
Explain what the area under a force-extension graph represents.
Describe how to find the work done from a curved force-extension graph.

2.3.6 Work done in stretching

Why the force changes during a stretch

The energy equation

Work done from a force-extension graph

Energy stored in a stretched spring

Energy from the area under a straight line

Extension from the energy stored

Work done by counting squares

Work done in stretching a rubber band

Work done in stretching

Review 2.3.6 Work done in stretching by teaching Genie

2.3.6 Work done in stretching

Why the force changes during a stretch

The energy equation

Work done from a force-extension graph

Energy stored in a stretched spring

Energy from the area under a straight line

Extension from the energy stored

Work done by counting squares

Work done in stretching a rubber band

Work done in stretching

Review 2.3.6 Work done in stretching by teaching Genie

2.3.6 Work done in stretching

Why the force changes during a stretch

The energy equation

Work done from a force-extension graph

Energy stored in a stretched spring

Energy from the area under a straight line

Extension from the energy stored

Work done by counting squares

Work done in stretching a rubber band

Work done in stretching

Review 2.3.6 Work done in stretching by teaching Genie

2.3.6 Work done in stretching Revision Guide

2.3.6 Work done in stretching

Why the force changes during a stretch

The energy equation

Work done from a force-extension graph

Energy stored in a stretched spring

Energy from the area under a straight line

Extension from the energy stored

Work done by counting squares

Work done in stretching a rubber band

Work done in stretching

Review 2.3.6 Work done in stretching by teaching Genie

2.3.6 Work done in stretching Revision Guide