5.3.1a Elastic and inelastic deformation

Forces that change an object’s shape

Definition

Deformation

Deformation is a change in the shape of an object caused by forces.

An object can be deformed by stretching, compressing or bending it.
1. Stretching: forces pull outwards on different parts, such as a spring stretched when its two ends are pulled apart.
2. Compressing: forces push inwards on different parts, such as a foam block squeezed between two hands.
3. Bending: forces act at different positions to curve the object, such as a ruler bent when its ends are supported and its centre pushed down.
The forces must act in suitable directions and at suitable positions; two forces pulling the same way would move the object rather than stretch it.

Why more than one force is needed

To change the shape of a stationary object, more than one force must be applied, because a stationary object has a resultant force of zero.
With only one force, the forces would be unbalanced and the object would accelerate instead of changing shape.
Two or more forces acting at different positions can be balanced overall while still changing the object’s shape.
1. For a spring stretched while stationary, one force pulls the left end left and another pulls the right end right.
2. The forces are equal and opposite, so the resultant is zero, but because they act on different ends they stretch the spring.
The same reasoning applies to compression: equal and opposite forces push inwards on opposite sides, keeping the object stationary while it becomes shorter.
In an answer, say the forces are balanced so the resultant is zero and the object stays stationary, and that they act on different parts so the object still deforms.

Common Mistake

Do not say that balanced forces cannot change an object; they do not change its motion, but they can change its shape if they act at different positions.
Do not ignore forces from supports or fixed points: if a person pulls one end of a spring while the other end is fixed to a wall, the wall exerts a force on that end.

Elastic deformation

Definition

Elastic deformation

Elastic deformation happens when an object returns to its original shape after the deforming forces are removed.

A stretched elastic band returns to its original length when released.
Its deformation was elastic because the change in shape was reversed once the stretching forces were removed.

Inelastic deformation

Definition

Inelastic deformation

Inelastic deformation happens when an object does not return completely to its original shape after the deforming forces are removed.

If a stretched object stays longer after the forces are removed, it has been inelastically deformed; the change in shape is not completely reversed.
The key difference is what happens after the forces are removed.
1. With elastic deformation, the object returns to its original shape.
2. With inelastic deformation, the object does not fully return to its original shape.
In a comparison, refer explicitly to removing the force: an elastically deformed object returns to its original shape, whereas an inelastically deformed object does not return completely.

Self review

What is deformation?
Give the forces involved in stretching and compressing an object.
Why must more than one force act to change the shape of a stationary object?
What happens to an elastically deformed object when the force is removed?
How does inelastic deformation differ from elastic deformation?

5.3.1b Hooke’s law and elastic potential energy (required practical)

Extension and Hooke’s law

Definition

Extension

The extension of an object is its increase in length: $length\text{extension} = \text{stretched length} - \text{original length}$ , measured in metres ( $m\text{m}$ ).

For many elastic objects, including springs, the extension is directly proportional to the applied force, provided the limit of proportionality is not exceeded.
This relationship is Hooke’s law: $F = k e$ .
1. $F$ is the force in newtons ( $N\text{N}$ ).
2. $k$ is the spring constant in newtons per metre ( $N/m\text{N/m}$ ).
3. $e$ is the extension in metres ( $m\text{m}$ ).
The spring constant measures stiffness: a larger $k$ means more force is needed for the same extension, so the spring is stiffer.
The equation also applies to compression, where $e$ is the compression rather than the extension.

Practical

Investigation: the relationship between force and extension of a spring

Aim: investigate how the extension of a spring depends on the force applied to it.
Apparatus: a spring with a loop at each end, a metre ruler, a splint and tape to act as a pointer, a 10 N weight stack added in 1 N steps, a clamp stand with two clamps and bosses, a G-clamp or heavy counterweight to stop the stand tipping, and safety goggles.
Set up the apparatus: clamp the spring so it hangs vertically; stand the metre ruler vertically beside it with the zero at the top of the spring; tape the splint pointer horizontally to the bottom of the spring so it rests against the ruler scale; clamp the stand to the bench, or add a counterweight, so it cannot tip.
Measure the unstretched length: read the pointer position on the ruler at eye level to avoid parallax and record it as the length with 0 N applied.
Add the first weight: hang the base of the weight stack on the spring to apply a force of 1.0 N (any mass in grams must first be converted to a weight in newtons using $W = m g$ ).
Read the new length: wait until the spring stops moving, read the pointer at eye level, and record the length for 1.0 N.
Repeat in 1 N steps up to about 10 N, recording the spring’s length each time, and stop before the spring is permanently stretched.
Calculate the extension for each force: $\text{stretched length} - \text{original length}$ .
Repeat and average: repeat the readings where possible and take a mean extension for each force to reduce random error; check and repeat any anomalous result rather than including it in a mean.
Plot a graph of extension (y-axis) against force (x-axis).
Variables: the independent variable is the force, the dependent variable is the extension, and control variables include using the same spring and the same reference point for each reading.
Result: a straight line through the origin shows the extension is directly proportional to the force (Hooke’s law); the point where the line begins to curve marks the limit of proportionality. The gradient equals $1k\dfrac{1}{k}$ , so a stiffer spring gives a shallower line.

Linear and non-linear behaviour

Before the limit of proportionality, a spring’s force–extension graph is a straight line through the origin, because force and extension are directly proportional.
A straight-line graph shows linear behaviour.
Beyond the limit of proportionality the graph curves: the relationship is non-linear, so doubling the force no longer doubles the extension, and $F = k e$ no longer holds with a constant $k$ .

Example

A spring extends by $m0.040\ \text{m}$ when a force of $N6.0\ \text{N}$ is applied. Calculate its spring constant.

Write the equation: $F = k e$ .
Rearrange for $k$ : $\dfrac{F}{e}$ .
Substitute the values: $\dfrac{6.0\ \text{N}}{0.040\ \text{m}}$ .
Calculate the spring constant: $150\ \text{N/m}$ .

Elastic potential energy

Definition

Elastic potential energy

Elastic potential energy is the energy stored in a stretched or compressed elastic object because work is done to deform it.

A force stretching or compressing a spring does work, transferring energy to its elastic potential energy store.
If the spring is not inelastically deformed, the work done equals the elastic potential energy stored.
Up to the limit of proportionality, $Ee=12ke2E_{\text{e}} = \tfrac{1}{2}ke^2$ $E_{e} = \frac{1}{2} k e^{2}$ .
1. $EeE_{\text{e}}$ is elastic potential energy in joules ( $J\text{J}$ ).
2. $k$ is the spring constant in newtons per metre ( $N/m\text{N/m}$ ).
3. $e$ is the extension (or compression) in metres ( $m\text{m}$ ).

Example

The same spring has a spring constant of $N/m150\ \text{N/m}$ and is extended by $m0.040\ \text{m}$ . Calculate the elastic potential energy stored.

Write the equation: $Ee=12ke2E_{\text{e}} = \tfrac{1}{2}ke^2$ .
Square the extension: $m2(0.040\ \text{m})^2 = 0.0016\ \text{m}^2$ .
Substitute the values: $m2E_{\text{e}} = 0.5 \times 150\ \text{N/m} \times 0.0016\ \text{m}^2$ .
Calculate the energy: $JE_{\text{e}} = 0.12\ \text{J}$ (the work done stretching the spring is also $J0.12\ \text{J}$ ).

Exam technique

Extension is not the total length; always subtract the original length.
Convert the extension into metres before using $F = k e$ or $Ee=12ke2E_{\text{e}} = \tfrac{1}{2}ke^2$ .
Hooke’s law applies only while force and extension are directly proportional.
When interpreting data, state the pattern and support it: the relationship is linear because equal increases in force give equal increases in extension, and the points form a straight line through the origin.
When finding energy, square the extension before multiplying by $12k\tfrac{1}{2}k$ .

Self review

State Hooke’s law and its equation.
What does a larger spring constant tell you about a spring?
How do you calculate the extension of a spring in the practical?
State the equation for elastic potential energy.
What does a curved force–extension graph show?

5.3.1a Elastic and inelastic deformation

Forces that change an object’s shape

Definition

Deformation

Deformation is a change in the shape of an object caused by forces.

An object can be deformed by stretching, compressing or bending it.
1. Stretching: forces pull outwards on different parts, such as a spring stretched when its two ends are pulled apart.
2. Compressing: forces push inwards on different parts, such as a foam block squeezed between two hands.
3. Bending: forces act at different positions to curve the object, such as a ruler bent when its ends are supported and its centre pushed down.
The forces must act in suitable directions and at suitable positions; two forces pulling the same way would move the object rather than stretch it.

Why more than one force is needed

To change the shape of a stationary object, more than one force must be applied, because a stationary object has a resultant force of zero.
With only one force, the forces would be unbalanced and the object would accelerate instead of changing shape.
Two or more forces acting at different positions can be balanced overall while still changing the object’s shape.
1. For a spring stretched while stationary, one force pulls the left end left and another pulls the right end right.
2. The forces are equal and opposite, so the resultant is zero, but because they act on different ends they stretch the spring.
The same reasoning applies to compression: equal and opposite forces push inwards on opposite sides, keeping the object stationary while it becomes shorter.
In an answer, say the forces are balanced so the resultant is zero and the object stays stationary, and that they act on different parts so the object still deforms.

Common Mistake

Do not say that balanced forces cannot change an object; they do not change its motion, but they can change its shape if they act at different positions.
Do not ignore forces from supports or fixed points: if a person pulls one end of a spring while the other end is fixed to a wall, the wall exerts a force on that end.

Elastic deformation

Definition

Elastic deformation

Elastic deformation happens when an object returns to its original shape after the deforming forces are removed.

A stretched elastic band returns to its original length when released.
Its deformation was elastic because the change in shape was reversed once the stretching forces were removed.

Inelastic deformation

Definition

Inelastic deformation

Inelastic deformation happens when an object does not return completely to its original shape after the deforming forces are removed.

If a stretched object stays longer after the forces are removed, it has been inelastically deformed; the change in shape is not completely reversed.
The key difference is what happens after the forces are removed.
1. With elastic deformation, the object returns to its original shape.
2. With inelastic deformation, the object does not fully return to its original shape.
In a comparison, refer explicitly to removing the force: an elastically deformed object returns to its original shape, whereas an inelastically deformed object does not return completely.

Self review

What is deformation?
Give the forces involved in stretching and compressing an object.
Why must more than one force act to change the shape of a stationary object?
What happens to an elastically deformed object when the force is removed?
How does inelastic deformation differ from elastic deformation?

5.3.1b Hooke’s law and elastic potential energy (required practical)

Extension and Hooke’s law

Definition

Extension

The extension of an object is its increase in length: $length\text{extension} = \text{stretched length} - \text{original length}$ , measured in metres ( $m\text{m}$ ).

For many elastic objects, including springs, the extension is directly proportional to the applied force, provided the limit of proportionality is not exceeded.
This relationship is Hooke’s law: $F = k e$ .
1. $F$ is the force in newtons ( $N\text{N}$ ).
2. $k$ is the spring constant in newtons per metre ( $N/m\text{N/m}$ ).
3. $e$ is the extension in metres ( $m\text{m}$ ).
The spring constant measures stiffness: a larger $k$ means more force is needed for the same extension, so the spring is stiffer.
The equation also applies to compression, where $e$ is the compression rather than the extension.

Practical

Investigation: the relationship between force and extension of a spring

Aim: investigate how the extension of a spring depends on the force applied to it.
Apparatus: a spring with a loop at each end, a metre ruler, a splint and tape to act as a pointer, a 10 N weight stack added in 1 N steps, a clamp stand with two clamps and bosses, a G-clamp or heavy counterweight to stop the stand tipping, and safety goggles.
Set up the apparatus: clamp the spring so it hangs vertically; stand the metre ruler vertically beside it with the zero at the top of the spring; tape the splint pointer horizontally to the bottom of the spring so it rests against the ruler scale; clamp the stand to the bench, or add a counterweight, so it cannot tip.
Measure the unstretched length: read the pointer position on the ruler at eye level to avoid parallax and record it as the length with 0 N applied.
Add the first weight: hang the base of the weight stack on the spring to apply a force of 1.0 N (any mass in grams must first be converted to a weight in newtons using $W = m g$ ).
Read the new length: wait until the spring stops moving, read the pointer at eye level, and record the length for 1.0 N.
Repeat in 1 N steps up to about 10 N, recording the spring’s length each time, and stop before the spring is permanently stretched.
Calculate the extension for each force: $\text{stretched length} - \text{original length}$ .
Repeat and average: repeat the readings where possible and take a mean extension for each force to reduce random error; check and repeat any anomalous result rather than including it in a mean.
Plot a graph of extension (y-axis) against force (x-axis).
Variables: the independent variable is the force, the dependent variable is the extension, and control variables include using the same spring and the same reference point for each reading.
Result: a straight line through the origin shows the extension is directly proportional to the force (Hooke’s law); the point where the line begins to curve marks the limit of proportionality. The gradient equals $1k\dfrac{1}{k}$ , so a stiffer spring gives a shallower line.

Linear and non-linear behaviour

Before the limit of proportionality, a spring’s force–extension graph is a straight line through the origin, because force and extension are directly proportional.
A straight-line graph shows linear behaviour.
Beyond the limit of proportionality the graph curves: the relationship is non-linear, so doubling the force no longer doubles the extension, and $F = k e$ no longer holds with a constant $k$ .

Example

A spring extends by $m0.040\ \text{m}$ when a force of $N6.0\ \text{N}$ is applied. Calculate its spring constant.

Write the equation: $F = k e$ .
Rearrange for $k$ : $\dfrac{F}{e}$ .
Substitute the values: $\dfrac{6.0\ \text{N}}{0.040\ \text{m}}$ .
Calculate the spring constant: $150\ \text{N/m}$ .

Elastic potential energy

Definition

Elastic potential energy

Elastic potential energy is the energy stored in a stretched or compressed elastic object because work is done to deform it.

A force stretching or compressing a spring does work, transferring energy to its elastic potential energy store.
If the spring is not inelastically deformed, the work done equals the elastic potential energy stored.
Up to the limit of proportionality, $Ee=12ke2E_{\text{e}} = \tfrac{1}{2}ke^2$ $E_{e} = \frac{1}{2} k e^{2}$ .
1. $EeE_{\text{e}}$ is elastic potential energy in joules ( $J\text{J}$ ).
2. $k$ is the spring constant in newtons per metre ( $N/m\text{N/m}$ ).
3. $e$ is the extension (or compression) in metres ( $m\text{m}$ ).

Example

The same spring has a spring constant of $N/m150\ \text{N/m}$ and is extended by $m0.040\ \text{m}$ . Calculate the elastic potential energy stored.

Write the equation: $Ee=12ke2E_{\text{e}} = \tfrac{1}{2}ke^2$ .
Square the extension: $m2(0.040\ \text{m})^2 = 0.0016\ \text{m}^2$ .
Substitute the values: $m2E_{\text{e}} = 0.5 \times 150\ \text{N/m} \times 0.0016\ \text{m}^2$ .
Calculate the energy: $JE_{\text{e}} = 0.12\ \text{J}$ (the work done stretching the spring is also $J0.12\ \text{J}$ ).

Exam technique

Extension is not the total length; always subtract the original length.
Convert the extension into metres before using $F = k e$ or $Ee=12ke2E_{\text{e}} = \tfrac{1}{2}ke^2$ .
Hooke’s law applies only while force and extension are directly proportional.
When interpreting data, state the pattern and support it: the relationship is linear because equal increases in force give equal increases in extension, and the points form a straight line through the origin.
When finding energy, square the extension before multiplying by $12k\tfrac{1}{2}k$ .

Self review

State Hooke’s law and its equation.
What does a larger spring constant tell you about a spring?
How do you calculate the extension of a spring in the practical?
State the equation for elastic potential energy.
What does a curved force–extension graph show?

5.3.1 Forces and elasticity

5.3.1a Elastic and inelastic deformation

Forces that change an object’s shape

Why more than one force is needed

Elastic deformation

Inelastic deformation

5.3.1b Hooke’s law and elastic potential energy (required practical)

Extension and Hooke’s law

Investigation: the relationship between force and extension of a spring

Linear and non-linear behaviour

Elastic potential energy

Recap questions

Review 5.3 Forces and elasticity by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.3.1 Forces and elasticity

5.3.1a Elastic and inelastic deformation

Forces that change an object’s shape

Why more than one force is needed

Elastic deformation

Inelastic deformation

5.3.1b Hooke’s law and elastic potential energy (required practical)

Extension and Hooke’s law

Investigation: the relationship between force and extension of a spring

Linear and non-linear behaviour

Elastic potential energy

Recap questions

Review 5.3 Forces and elasticity by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.3.1 Forces and elasticity

5.3.1a Elastic and inelastic deformation

Forces that change an object’s shape

Why more than one force is needed

Elastic deformation

Inelastic deformation

5.3.1b Hooke’s law and elastic potential energy (required practical)

Extension and Hooke’s law

Investigation: the relationship between force and extension of a spring

Linear and non-linear behaviour

Elastic potential energy

Recap questions

Review 5.3 Forces and elasticity by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.3 Forces and elasticity Revision Guide

5.3.1 Forces and elasticity

5.3.1a Elastic and inelastic deformation

Forces that change an object’s shape

Why more than one force is needed

Elastic deformation

Inelastic deformation

5.3.1b Hooke’s law and elastic potential energy (required practical)

Extension and Hooke’s law

Investigation: the relationship between force and extension of a spring

Linear and non-linear behaviour

Elastic potential energy

Recap questions

Review 5.3 Forces and elasticity by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.3 Forces and elasticity Revision Guide