Back to course
8463 · RP6; 8.2.6Required practical activitiesCore

RP6: Spring Force and Extension

Secure the spring and place a ruler alongside it using an approved stable arrangement. Record its original length or the initial position of a fixed pointer. Add suitable loads in steps, let oscillations settle and record the corresponding positions. Extension is the increase from the original reference, not the spring’s total loaded length. Read the scale at eye level and use the same pointer each time.

35 min Video by Malmesbury Education Required practical activities
Hooke's Law - GCSE Science Required PracticalCurated by Kepler, not produced by us. Watch it before the notes below.
Open on YouTube

Before you start

  • Use your teacher’s tier checklist: this overview can include Higher-only material, and is not a tier-specific revision list.
  • AQA separate GCSE Physics 8463. Use the specified required-practical reference and the centre’s approved apparatus and supervised procedure.
  • Distinguish mass and weight, convert centimetres to metres and find a graph gradient.
  • Use the centre’s approved supervised method and equipment limits. These notes prepare you to explain and evaluate practical work; they do not replace carrying it out and recording the evidence.

Watch, pause, try

Video by Malmesbury Education. The demonstration develops force and extension. Pearson’s written guide additionally requires work done, which is developed through graph area and the original examples.

Viewing focus: Calculate extension from loaded and original lengths

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: A spring length changes from 11 cm to 15 cm. Which value is x, and what is it in metres?

Show answer and reasoning

The extension is 4 cm = 0.040 m, not the total 15 cm loaded length.

After watching 2: How would the meaning of a straight-line gradient change if the force and extension axes were exchanged?

Show answer and reasoning

Force against extension gives k in N m⁻¹. Extension against force gives its reciprocal, 1/k in m N⁻¹.

What you'll be able to do

  • Calculate extension from loaded and original lengths
  • Identify a proportional force–extension region
  • Calculate spring constant and work with stated elastic assumptions
1

Measure extension from one reference

Secure the spring and place a ruler alongside it using an approved stable arrangement. Record its original length or the initial position of a fixed pointer. Add suitable loads in steps, let oscillations settle and record the corresponding positions. Extension is the increase from the original reference, not the spring’s total loaded length. Read the scale at eye level and use the same pointer each time.

Convert each added mass to its weight using the supplied gravitational field strength. In a static hanging arrangement, the spring’s upward force balances the load’s downward weight. A 100 g mass is not a 100 N force. Keep within approved limits, secure the stand and use the centre’s arrangements to contain falling loads. A loading and unloading comparison can reveal whether the spring returns to its initial length.

Two schematic spring lengths share a fixed upper reference. The original length is 12 centimetres and loaded length 16 centimetres; the lower-end displacement is 4 centimetres, equal to 0.040 metres.
Scroll the diagram to see the full view.
Original measurement schematic. Extension is a change from the unloaded reference; it is not the total loaded length. The load is stationary for each reading.
2

Read proportionality from the chosen axes

Within a proportional region, F = kx: force increases in proportion to extension and a force-against-extension graph is straight through the appropriate zero reference. Its gradient is k in N m⁻¹. An extension-against-force graph instead has gradient 1/k. State the axes before interpreting a slope. A straight line with a substantial unexplained intercept needs its zero reference or model checked.

Beyond the limit of proportionality, force and extension no longer have a constant ratio. That limit is not defined as identical to the elastic limit: a non-proportional deformation may still be recoverable. Inspect the data rather than extending an early straight line through every later point. Record unloading evidence before making a claim about permanent deformation.

Extension / mSpring A: force / NSpring B: force / N
0.0000.00.0
0.0100.51.0
0.0201.02.0
0.0301.53.0
0.0402.04.0
0.0502.55.0
0.0603.06.0
Scroll horizontally to see the full code or table.
Original illustrative proportional-region data for two different springs. These are model comparisons, not measurements from a single spring or evidence beyond the displayed range.

Compare spring stiffness from force and extension

Calculate the force–extension gradient for each spring. Which needs more force for the same extension, and why is total spring length not the horizontal quantity?

Spring A: Force / N against Extension / mData points are joined by straight guides. A dashed line joins the two selected measurements. Use the selectors below to read values and compare intervals; this line gives an average gradient, not an instantaneous tangent.302.250.0151.50.030.750.04500.06Extension / mForce / N

Blue: data with straight joining guides. Gold: selected interval. Scroll the plot horizontally on a small screen.

Try Δy ÷ Δx on paper first. Keep the sign and use the units on the axes.

The gradient gives k because force is plotted against extension. The straight guides describe the stated proportional model; they do not establish the limit of proportionality of a real spring.
3

Connect the area to transferred energy

Work done in slowly stretching a spring is represented by the area beneath a force-against-extension graph. In the proportional region the area is a triangle, so E = ½Fx = ½kx². The final force multiplied by the entire extension would be too large because the force increased during the stretch. Use metres for x to obtain joules.

Interpreting the work as recoverable elastic energy assumes the stated elastic model. If the spring is permanently deformed or shows substantial hysteresis, not all loading work is recovered. For a non-linear curve, the triangular shortcut is not generally valid. Estimate the actual area only when the question supplies a suitable method or enough information.

4

Worked applications

Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.

Example 1A spring’s original length is 12.0 cm and its loaded length is 16.0 cm under a 2.00 N force. Find extension and k within the proportional region.
1Extension = 16.0 − 12.0 = 4.0 cm = 0.040 m.
2k = F/x = 2.00/0.040.
Answer
Extension 0.040 m; spring constant 50 N m⁻¹.
Example 2A proportional spring with k = 50 N m⁻¹ is extended by 0.060 m. Calculate the work done.
1E = ½kx².
2E = 0.5 × 50 × 0.060².
Answer
0.090 J, assuming the stated proportional elastic model.
Example 3An extension-against-force graph rises by 0.040 m when force rises by 2.00 N. Find its gradient and spring constant.
1Graph gradient = Δx/ΔF = 0.040/2.00 = 0.020 m N⁻¹.
2This slope is 1/k, so k = 1/0.020.
Answer
Gradient 0.020 m N⁻¹; k = 50 N m⁻¹.

Common mistakes to avoid

Putting centimetres directly into a formula using N m⁻¹.
Convert extension to metres.
Equating non-proportionality automatically with permanent deformation.
Use unloading evidence to discuss elastic recovery.

Key takeaways

  • You can calculate extension from loaded and original lengths.
  • You can identify a proportional force–extension region.
  • You can calculate spring constant and work with stated elastic assumptions.

Apply it in an exam

Question 1: An extension-against-force graph rises by 0.040 m when force rises by 2.00 N. Find its gradient and spring constant.

Show answer and reasoning

Gradient 0.020 m N⁻¹; k = 50 N m⁻¹.

  • Graph gradient = Δx/ΔF = 0.040/2.00 = 0.020 m N⁻¹.
  • This slope is 1/k, so k = 1/0.020.

Explore further

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

  • Which slope gives k directly?
  • Why is proportional-region work ½Fx?

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 10 lessons