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P10.5 Forces and Elasticity

GCSE Combined Science · AQA · Chapter P10: Force and Motion

A free GCSE Combined Science: Trilogy lesson for AQA (8464) from OBA School: elastic and inelastic deformation, F = k e, the limit of proportionality and the work done in stretching a spring. 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 for stretching and compressing, converting mm or cm to m
  • Interpret force–extension graphs: linear and non-linear parts, the spring constant from the gradient and 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.
  • Extension: Stretched length − original (unstretched) length.
  • 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 or bending 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 gently squeezed drinks can, a hair bobble, a diving board.
  • Inelastic deformation: the object stays changed – a crushed drinks can, a bent paper clip, a spring stretched too far.

Extension and F = k e

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. It also works for compression (e is then the decrease in length).

Worked example: finding the spring constant

A spring is 6.0 cm long. With a 4.5 N weight on it, it is 9.0 cm long.
e = 9.0 − 6.0 = 3.0 cm = 0.030 m; k = F ÷ e = 4.5 ÷ 0.030

Answer: k = 150 N/m

Worked example: finding the extension

A spring with k = 80 N/m supports a 3.2 N weight.
e = F ÷ k = 3.2 ÷ 80 = 0.040 m

Answer: e = 0.040 m (4.0 cm)

Worked example: a compressed spring

A force of 750 N compresses a car suspension spring by 2.5 cm.
k = F ÷ e = 750 ÷ 0.025

Answer: k = 30 000 N/m – a very stiff spring

Exam tip: Change mm or cm into metres before using F = k e: 35 mm = 0.035 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 the extension is directly proportional to the force. 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). A spring stretched this far is usually deformed inelastically and does not go back to its original length. Some materials, such as rubber bands, give a non-linear graph even though they deform elastically.

Work done and 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 toy launcher). If an object is deformed inelastically, the work done warms it slightly instead.

elastic potential energy = 0.5 × 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 = 60 N/m is stretched by 0.15 m.
Ee = ½ × 60 × 0.15² = ½ × 60 × 0.0225

Answer: Ee = 0.675 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.
  • Applying F = k e 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 45 N/m. Calculate the force needed to stretch it by 0.20 m.

Answer: 9 N

Working: F = 45 × 0.20 = 9.0 N.

Question 2. A force of 6.0 N stretches a spring by 4.0 cm. Calculate the spring constant.

Answer: 150 N/m

Working: e = 0.040 m. k = 6.0 ÷ 0.040 = 150 N/m.

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

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

Working: It is the point where the graph stops being a straight line: beyond it, extension is no longer proportional to force.

Related lessons

Continue in the full course

This lesson is section P10.5 (AGCSC3905) of our free GCSE Combined Science 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 Combined Science course.

 

 

 

 

 

 

 

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