10.8 Forces and Elasticity

A free GCSE Physics lesson for AQA (8463) from OBA School: elastic and inelastic deformation, Hooke’s law F = k e, the limit of proportionality and elastic potential energy. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

• Explain why more than one force is needed to change the shape of an object, and tell elastic deformation from inelastic deformation
• Calculate extension and use F = k e, rearranging it and converting mm or cm to m
• Interpret force–extension graphs: find the spring constant from the gradient and identify the limit of proportionality
• Calculate the work done in stretching a spring (the elastic potential energy stored) using Ee = ½ k e²

Key words

• Elastic deformation: The object returns to its original shape and size when the forces are removed.
• Inelastic deformation: The object does not return to its original shape when the forces are removed (also called plastic deformation).
• Extension: Stretched length − original length.
• Hooke’s law: The extension of a spring is directly proportional to the force applied, up to the limit of proportionality.
• Spring constant, k: The force needed per metre of extension, in N/m. A stiffer spring has a bigger k.
• Limit of propor- tionality: The point beyond which extension is no longer directly proportional to force.
• Linear / non-linear: A linear relationship gives a straight-line graph; a non-linear one gives a curve.
• Elastic potential energy: Energy stored in a stretched or squashed elastic object, in joules.

Changing shape

You can change the shape of an object by stretching, compressing, bending or twisting it. This always needs more than one force. With only one force the object would simply accelerate. A spring hanging from a clamp is stretched by the weight pulling down and the clamp pulling up.

• Elastic deformation: the object goes back to its original shape when the forces are removed – a trampoline spring, a rubber band, a squash ball.
• Inelastic deformation: the object stays changed – modelling clay, a crushed drinks can, an over-stretched polythene bag.

Extension and Hooke’s law

extension = stretched length − original length

force = spring constant × extension F = k e

F = force in newtons, N; k = spring constant in newtons per metre, N/m; e = extension in metres, m

Rearranged: k = F ÷ e and e = F ÷ k. Hooke’s law also works for compression (e is then the decrease in length).

Worked example: finding the spring constant

A spring is 8.0 cm long. With a 6.0 N weight on it, it is 11.0 cm long.
e = 11.0 − 8.0 = 3.0 cm = 0.030 m
k = F ÷ e = 6.0 ÷ 0.030

Answer: k = 200 N/m

Worked example: finding the extension

A spring with k = 40 N/m supports a 2.4 N weight.
e = F ÷ k = 2.4 ÷ 40 = 0.060 m

Answer: e = 0.060 m (6.0 cm)

Exam tip: Change mm or cm into metres before using F = k e: 25 mm = 0.025 m; 4.0 cm = 0.040 m.

Force–extension graphs

Plot force on the y-axis and extension on the x-axis. For a spring the graph is a straight line through the origin – a linear relationship – so it obeys Hooke’s law. The gradient of the straight part is the spring constant k.

Past the limit of proportionality the line curves: the extension is no longer proportional to the force (non-linear). If the spring is stretched further still, past its elastic limit, it will not go back to its original length.

Elastic potential energy

When a spring is stretched, the force does work on it. If the spring is not inelastically deformed, all of this work is stored as elastic potential energy, and it can be transferred back (as in a catapult or a bow).

elastic potential energy = ½ × spring constant × (extension)² Ee = ½ k e²

Ee in joules, J; k in N/m; e in m. Use it only up to the limit of proportionality.

Worked example: energy in a stretched spring

A spring with k = 50 N/m is stretched by 0.12 m.
Ee = ½ × 50 × 0.12² = ½ × 50 × 0.0144

Answer: Ee = 0.36 J

Common mistakes

• Using the stretched length instead of the extension in F = k e.
• Leaving the extension in mm or cm in F = k e or Ee = ½ k e².
• Forgetting to square the extension in Ee = ½ k e².
• Plotting extension on the y-axis and calling the gradient k. With extension on the y-axis the gradient is 1/k.
• Applying Hooke’s law beyond the limit of proportionality.

Revision questions

Try these, then check your answers below each one.

Question 1. A spring has a spring constant of 30 N/m. Calculate the force needed to stretch it by 0.20 m.

Answer: 6 N

Working: F = 30 N/m × 0.20 m = 6.0 N.

Question 2. A force of 5.0 N stretches a spring by 2.5 cm. Calculate the spring constant.

Answer: 200 N/m

Working: e = 2.5 cm = 0.025 m. k = 5.0 ÷ 0.025 = 200 N/m.

Question 3. What is meant by the limit of proportionality of a spring? (Choose from: The largest force a spring can take before it breaks, The force needed to stretch the spring by 1 m, The original length of the spring, The point beyond which extension is no longer directly proportional to force.)

Answer: The point beyond which extension is no longer directly proportional to force

Working: Up to the limit of proportionality the force–extension graph is a straight line through the origin. Beyond it the graph curves (non-linear).

Related lessons

• Previous lesson: 10.7 Safety First
• Next lesson: 10.9 Required Practicals: Springs and Acceleration
• In this chapter: 10.1 Force and Acceleration
• In this chapter: 10.2 Weight and Terminal Velocity
• In this chapter: 10.3 Forces and Braking

Continue in the full course

This lesson is section 10.8 (AGCPH1008) of our free GCSE Physics course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Physics course.