Forces & Elasticity Notes

Edexcel GCSE Physics: Revision notes

Key facts

Elastic distortion: the object returns to its original shape. Inelastic distortion: it does not.
Hooke's law: $F = k \times e$ , with extension proportional to force up to the limit of proportionality.
Extension = new length − natural length.
The gradient of a force–extension graph is the spring constant $k$ (N/m).
$E_e = \tfrac{1}{2} k e^2$ is the energy stored in a stretched spring (Higher tier).

Hooke's law $F = k \times e$
Elastic energy $E_e = \tfrac{1}{2} k e^2$

Elastic and inelastic distortion

A force can stretch, compress or bend an object; if it springs back it is elastic, and if it stays deformed it is inelastic.

When a force is applied, an object can stretch, compress or bend: this is distortion.

With elastic distortion the object returns to its original shape and length once the force is removed. With inelastic distortion it does not: it has been permanently deformed. A spring stretched too far is damaged and shows inelastic distortion.

Force–extension for a spring (illustrative values): elastic and proportional up to the limit, then extension grows faster than the force

Elastic

Returns to original shape and length
Example: rubber band stretched a little

Inelastic

Stays permanently deformed
Example: rubber band stretched until baggy

A spring is stretched and, when the force is removed, stays longer than before. This is:

Force and extension

Up to the limit of proportionality, a spring's extension is directly proportional to the force applied.

Extension is the increase in length from the natural length. For a spring, force $F$ (N) = spring constant $k$ (N/m) × extension $e$ (m). This applies to stretching and compression.

The graph is a straight line until the limit of proportionality, then it curves. The elastic limit is the point beyond which the spring no longer returns to its original length.

k = 40spring constant (N/m)

Hooke's law: change k to see how stiffness changes the gradient

A spring is 12 cm long unstretched and 15 cm long under a load. Its extension is:

The spring constant

The spring constant is the force needed per metre of extension: a large value means a stiff spring.

The spring constant $k$ is in N/m. A large $k$ means a stiff spring and a small $k$ a floppy one. Rearranging gives $k = F \div e$ , which is the gradient of the straight part of a force–extension graph.

k = 20spring constant (N/m)

F = ke: the gradient of a force–extension line is the spring constant k (use the slider); a steeper line is a stiffer spring

Spring constant (N/m) $k = \dfrac{F}{e}$

Worked example

A force of 6 N stretches a spring by 0.15 m. Calculate the spring constant.

1
Write the equation: $F = k \times e$ .
2
Rearrange: $k = F \div e$ .
3
Substitute: $k = 6 \div 0.15$ .

Spring A has k = 20 N/m and spring B has k = 60 N/m. Which is stiffer?

Core practical: extension of a spring

Add masses in equal steps, measure the extension each time, then plot force against extension.

The graph is a straight line through the origin while the spring is proportional, and its gradient equals $k$ . The area under the graph gives the work done in stretching the spring.

1
Natural length
Measure with a ruler before adding any mass.
2
Add a known mass
Measure the new length.
3
Calculate extension
New length − natural length.
4
Repeat
In equal steps, up to the elastic limit.
5
Plot
Force against extension: gradient = k; area = work done.

Investigating extension of a spring

In this practical, what does the gradient of the straight part of the graph give?

Elastic potential energy

A spring stretched elastically stores energy equal to half the spring constant times the extension squared.

Stretching a spring elastically does work on it, stored as elastic potential energy ( $E_e$ in joules). Beyond the limit of proportionality, the work done is greater than the energy stored, because some is used to deform the spring permanently, for example as heat.

Shaded area = energy stored: ½ × 0.2 × 8 = 0.8 J

Elastic potential energy (J) $E_e = \tfrac{1}{2} k e^2$

Worked example

A spring with k = 40 N/m is stretched by 0.2 m. How much energy does it store?

1
Write the equation: $E_e = \tfrac{1}{2} k e^2$ .
2
Square first: $0.2^2 = 0.04$ .
3
Substitute: $E_e = 0.5 \times 40 \times 0.04$ .

The extension of a spring doubles. The energy stored becomes:

Try an exam question

A spring has a spring constant of 50 N/m. A force of 4 N stretches it elastically. Calculate the extension, and the energy stored in the spring.

[4 marks]

That's the notes covered.

Carry on to the next subtopic.