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Physics · Forces
Forces and elasticity
Describe elastic and inelastic deformation, use \(F = ke\) and \(E_e = \tfrac{1}{2}ke^2\), and investigate the relationship between force and extension (Required Practical 6).
Teacher resources
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- Forces and elasticity - Teacher Slides.pptx Teacher The lesson slides with the teacher's notes on each slide, and every question and mark scheme built in. Built from the lesson script on 30 September 2026. View
- Forces and elasticity - Teacher Notes.docx Teacher The complete notes with the teacher's notes and every model answer in full. Built from the lesson script on 30 September 2026. View
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- Forces and elasticity.pptx Built from the lesson script on 30 September 2026. View
- Forces and elasticity - Completed Notes.docx The full notes for the lesson, to revise from. Built from the lesson script on 30 September 2026. View
- Forces and elasticity - Exam Questions.docx Built from the lesson script on 30 September 2026. View
Warm-up
Answer each one, then check.
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1
What is a spring?
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A coiled wire that can be stretched or compressed
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2
What does proportional mean?
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Doubling one doubles the other
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3
Convert 6 cm to m.
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0.06 m
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4
What is elastic potential energy?
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Energy stored in a stretched spring
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5
What is the unit of force?
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Newton
Learning Objectives
- 1Explain why more than one force is needed to stretch, bend or compress an object.
- 2Describe elastic and inelastic deformation.
- 3Recall and apply \(F = ke\) and \(E_e = \tfrac{1}{2}ke^2\).
- 4Interpret force-extension graphs and describe Required Practical 6.
HOOKE'S LAW
force \(=\) spring constant \(\times\) extension \(F = ke\)
The extension of an elastic object is directly proportional to the force applied, provided the limit of proportionality is not exceeded. The same applies to compression.
Force and Extension
Beyond the limit, the spring does not return to its original length.
Required Practical 6: Stretching a Spring
Measure the length with each extra mass added.
Required Practical 6: Method
Investigate force and extension.
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1
Measure the original length
With no masses on the spring.
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2
Add a mass
Record the force (weight = mg) and the new length.
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3
Calculate the extension
New length minus original length.
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4
Repeat
Add masses one at a time, keeping below the limit of proportionality at first.
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5
Plot
Force (y-axis) against extension (x-axis).
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6
Gradient
Gradient of the straight part = spring constant k.
Spring Constant
A force of 3.0 N stretches a spring by 6.0 cm (within the limit of proportionality). Calculate the spring constant.
Show the solutionHide the solution
- 1 Convert \(6.0\) cm \(= 0.060\) m
- 2 Rearrange \(k = F \div e\)
- 3 Substitute \(k = 3.0 \div 0.060\)
- 4 Answer \(k = 50\) N/m
Answer50 N/m
Elastic Potential Energy
Calculate the energy stored in the spring when the extension is 0.040 m and k = 50 N/m.
Show the solutionHide the solution
- 1 Write the equation \(E_e = \tfrac{1}{2}ke^2\)
- 2 Substitute \(E_e = 0.5 \times 50 \times 0.040^2\)
- 3 Answer \(E_e = 0.040\) J
Answer0.040 J
Elastic or Inelastic?
Elastic deformation
- The object returns to its original shape when the force is removed.
- Happens within the limit of proportionality.
Inelastic deformation
- The object does not return to its original shape.
- Happens beyond the limit of proportionality.
Key Ideas
Learn these.
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Two forces
To stretch, bend or compress a stationary object, more than one force is needed (for example pull at both ends).
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Work done
The work done on an elastic object equals the elastic potential energy stored, provided it is not inelastically deformed.
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Linear
A straight line through the origin: proportional.
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Non-linear
A curve: not proportional.
Spring Test
A spring stretches 4.0 cm when a 2.0 N force is applied. (a) Calculate k in N/m. (b) What force gives an extension of 10 cm if the limit is not exceeded? (c) What energy is stored at 10 cm?
1. Convert cm to m.
2. Use F = ke and Ee = ½ke².
A good answer shows: (a) 50 N/m. (b) F = 50 × 0.10 = 5.0 N. (c) E = 0.5 × 50 × 0.10² = 0.25 J.
Can I...?
- 1State F = ke.
- 2Convert cm to m.
- 3Calculate a spring constant.
- 4Calculate stored energy.
- 5Describe elastic and inelastic deformation.
- 6Read a force-extension graph.
- 7Describe Required Practical 6.
- 8Find the gradient.
Summary & Exam Focus
- \(F = ke\).
- \(E_e = \tfrac{1}{2}ke^2\).
- Beyond the limit of proportionality: non-linear, may not return.
- Spring constant = gradient of the linear part.
Exam focus
A spring has a spring constant of 50 N/m. Calculate the force needed to stretch it by 0.10 m. (2 marks) (2 marks)
Use F = ke.
Key terms
The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.
- Extension
- The increase in length of a spring.
- Spring constant
- The force needed per metre of extension (N/m).
- Limit of proportionality
- The point beyond which force and extension are no longer proportional.
- Elastic deformation
- A change of shape that is reversed when the force is removed.
- Inelastic deformation
- A permanent change of shape.
- Linear
- Following a straight line.
Questions and answers
10 questions set on this lesson, with the mark schemes and model answers open.
A spring has a spring constant of 50 N/m. Calculate the force needed to stretch the spring by 0.10 m. Use the equation: force = spring constant × extension
Mark scheme — 2 marks available
- Correct substitution — 1 mark
- 5.0 N — 1 mark
Model answer
\(50 \times 0.10 = 5.0\) N
A student investigates a spring. The graph shows how the force changes with extension. (a) State the extension at the limit of proportionality. (b) Calculate the spring constant of the spring. (c) Calculate the elastic potential energy stored when the extension is 0.040 m.
Mark scheme — 5 marks available
- 6 cm (accept 5.5 to 6.5) — 1 mark
- Uses 3.0 N at 0.060 m — 1 mark
- 50 N/m — 1 mark
- Correct substitution — 1 mark
- 0.040 J — 1 mark
Model answer
(a) About 6 cm. (b) \(k = 3.0 \div 0.060 = 50\) N/m. (c) \(E_e = 0.5 \times 50 \times 0.040^2 = 0.040\) J
Describe the difference between elastic deformation and inelastic deformation of a spring.
Mark scheme — 2 marks available
- Elastic: returns to original shape — 1 mark
- Inelastic: does not return — 1 mark
Model answer
In elastic deformation the spring returns to its original length when the force is removed; in inelastic deformation it does not and is permanently changed.
Describe how to investigate the relationship between the force applied to a spring and the extension of the spring. Your answer should include the apparatus, the measurements, and how you would use them.
Mark scheme — 6 marks available
- Level 3 (5 to 6 marks): a complete method with stand, ruler and masses, original length measured, several forces (weights) applied with the new length each time, extension calculated, graph plotted and spring constant from the gradient, with a precaution (eye level, avoid overstretching) — 5 to 6 marks
- Level 2 (3 to 4 marks): a method covering force and extension with some detail missing — 3 to 4 marks
- Level 1 (1 to 2 marks): simple statements about adding masses and measuring length — 1 to 2 marks
Model answer
See levels-of-response scheme.
Explain why two forces are needed to stretch a spring that is fixed to a wall.
Mark scheme — 2 marks available
- Forces act at both ends — 1 mark
- Equal and opposite forces — 1 mark
Model answer
The wall applies a force in the opposite direction to the pulling force, so the spring is stretched; with only one force the object would accelerate rather than change shape.
Extension is...
Why: It is the increase in length.
A 4 N force gives a 0.08 m extension. k =
Why: 4 ÷ 0.08 = 50.
A straight line force-extension graph through the origin shows...
Why: F ∝ e.
Beyond the limit of proportionality the spring...
Why: It is inelastically deformed.
The gradient of the linear section is the...
Why: F ÷ e.