Core practical: stretching a spring
Core practical 15.6 investigates how the extension of a spring depends on the force applied, and how much energy it stores.
The method
2. Fix a pointer to the bottom of the spring and record its position (the original length).
3. Add a 1 N weight, record the new position and work out the extension (new length − original length).
4. Add weights one at a time up to about 6 N. Read the ruler at eye level to avoid parallax error.
5. Plot force (y-axis) against extension (x-axis).
Safety: wear eye protection in case the spring snaps, and keep feet clear of falling weights.
Reading the graph
The gradient of the line is the spring constant, k, in N/m.
Beyond the limit of proportionality, the line curves (a non-linear relationship). If stretched too far, the spring will not return to its original length (inelastic deformation).
Work done and energy stored
E = ½ × k × x2 (only up to the limit of proportionality).
This is the same as the area under the force-extension graph: E = ½ × F × x.
A spring's length goes from 0.150 m to 0.190 m when a 2.0 N weight is hung on it. Find the spring constant and the energy stored.
- x = 0.190 − 0.150 = 0.040 m
- k = F ÷ x = 2.0 ÷ 0.040 = 50 N/m
- E = ½ × 50 × 0.0402 = 0.04 J
Answer: k = 50 N/m; energy stored = 0.04 J
Measure the extension for each added weight with a pointer and ruler at eye level. A straight line through the origin shows F = kx; the gradient is k. Beyond the limit of proportionality the graph curves. Energy stored = ½kx2, the area under the graph.
- Extension
- The increase in length of a spring or wire when it is stretched.
- Weight
- The force of gravity on an object: W = m × g. Unit newton (N).
- Limit of proportionality
- The point beyond which extension is no longer proportional to the force.
- Time
- 1 hour = 3600 s. Convert minutes and hours to seconds before using P = E ÷ t or Q = I × t.
- Inelastic deformation
- A change of shape that stays after the force is removed.
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2 more questions
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A spring is 10.0 cm long. With a weight on it, it is 13.2 cm long. What is the extension in metres?
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Answer: 0.032 m
13.2 − 10.0 = 3.2 cm = 0.032 m
A force of 2.0 N gives an extension of 0.050 m. Calculate the spring constant.
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Answer: 40 N/m
k = F ÷ e = 2.0 ÷ 0.050 = 40 N/m
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- Pearson Edexcel GCSE Physics (1PH0) specification, topic 15.3, 15.4, 15.5, 15.6
- Pearson Edexcel International GCSE Physics (4PH1) specification, topic 1.22, 1.23, 1.24
- Pearson Edexcel GCSE (9-1) Physics (1PH0) specification
Checked by the GCSE-ready editorial team, 17 October 2026. Last updated 2 October 2026. Spotted a mistake? Tell us and we will fix it.
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