Lesson focus: Spring extensionHow much longer a spring or elastic band gets when it is stretched. can be proportional to forceA push or a pull, measured in newtons (N). without elasticity being guaranteed.
Calculate extension, analyse force–extension data and distinguish two limits.
Build on earlier learning
Squashing and stretchingQuantify stiffness and test recovery after deformation.
You will be successful when you can…
- Calculate extension.
- Plot force against extension.
- Identify proportional region.
- Distinguish proportional and elastic limits.
Words that carry the science
- extension
- Loaded length minus original length.
- Hooke’s law
- Force is proportional to extension within the proportional region.
- limit of proportionality
- Point beyond which force and extension are no longer proportional.
- elastic limit
- Greatest load after which full recovery still occurs.
- spring constant
- Force per unit extension, N/m.
Measure extension before interpreting graph shape
01Calculate the change
A spring from 8.0 cm to 11.0 cm has 3.0 cm extensionHow much longer a spring or elastic band gets when it is stretched., not 11.0 cm. Read rulers at eye level.
02Read the graph
A straight line through origin means force/extension ratio is constant. Curvature marks loss of proportionality.
Hooke’s lawThe extension of a spring is proportional to the force on it, so doubling the force doubles the extension, up to the limit of proportionality. says the extension doubles when the force doubles. To check this, you need several readings at different forces, plotted on a graph. One reading on its own cannot show a pattern.
03Unload to test elasticity
Graph proportionality does not by itself show whether the spring returns. Unloading evidence locates permanent deformation.
Stretching a spring
Add weights one newton at a time and watch the extension. Remove them all to test whether the spring returns to its original length.
Controls
Illustrative spring: 4 cm per newton up to 5 N; it stays stretched if loaded beyond 6 N.
Calculate stiffness
A 4.0 N force produces 2.0 cm extension in the proportional region.
- Convert 2.0 cm to 0.020 m.
- Use k = F ÷ x.
- k = 4.0 ÷ 0.020.
- k = 200 N/m.
How to make the reasoning visible: Unit conversion is an IGCSE bridge; the graph reasoning remains the core.
Spring dataset
Original length is 8.0 cm.
| Force / N | Length / cm | Extension / cm | After unloading / cm |
|---|---|---|---|
| 0 | 8.0 | 0.0 | 8.0 |
| 1 | 9.0 | 1.0 | 8.0 |
| 2 | 10.0 | 2.0 | 8.0 |
| 3 | 11.2 | 3.2 | 8.0 |
| 4 | 13.0 | 5.0 | 8.6 |
Graph skills: SLAPUK and choosing a scaleSpring data: when to draw a ruled line of best fit.
Where does proportionality begin to fail?
What shows permanent deformation?
Use the dataset. What is the extension when the force is 2 N?
Check your understanding
Answer from memory, then check against the suggested answer.
What is extension?
What is Hooke’s law?
What is limit of proportionality?
How do you calculate extension?
What is the elastic limit?
What error results from viewing a ruler at an angle?
A second spring has k = 400 N/m. Compare stiffness.
Up to 2 N, each extra 1 N stretches the spring by 1 cm. Predict the extension for 1.5 N, and say why you cannot use the same rule at 4 N.
Exam Question Practice
Exam Question Practice
Mark points shown after marking
A spring is 7.5 cm long. With a 3 N weight hanging from it, it is 10.8 cm long.
Calculate the extension of the spring.
(2)A student says this one reading proves Hooke’s law. Explain whether they are right.
(2)Total for this question: 4 marks
Command word: Calculate first, then evaluate what the evidence can establish.