GCSE Physics resources
AQA GCSE Forces
Extension is the change in length. A force–extension graph tells you whether proportionality holds, while unloading tells you whether deformation is reversible.
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Extension is the change in length. A force–extension graph tells you whether proportionality holds, while unloading tells you whether deformation is reversible.
Before you revise
Choose an answer from memory. Your result tells you what to watch for in the guide.
Core revision guide
Extension is the change in length. A force–extension graph tells you whether proportionality holds, while unloading tells you whether deformation is reversible.
Scope: 8463 and 8464 · Foundation and Higher
Measure e = loaded length − unloaded length. Within the proportional region F = ke, so the gradient of F (y) against e (x) is k in N/m. Convert centimetres to metres before computing k. A larger k means more force for the same extension, describing a stiffer spring.
Scope: 8463 and 8464 · Foundation and Higher
The limit of proportionality is where F is no longer proportional to e and the line begins to curve. Elastic deformation is reversed when the force is removed; inelastic deformation leaves a permanent change. A curved graph alone does not establish that the spring is permanently deformed: check the unloaded length.
Scope: 8463 and 8464 · Foundation and Higher
Two opposing forces are needed to stretch a stationary object. Work done in stretching raises its elastic potential store. In the proportional region, Ee = ½ke² = ½Fe, equal to area under F–e. Force rises from zero to F, so using Fe would double the work for this linear case.
Misconception clinic
Read each belief, say what is wrong with it, then compare your correction.
Tempting idea: 13 m.
Use instead: Neither subtraction nor unit conversion is correct. (13 − 8) cm = 0.05 m.
Tempting idea: The spring must be permanently stretched.
Use instead: Permanent deformation requires an unloading check. The relationship is no longer proportional.
Retrieval practice
Explore the relationship
Change the force or extension, then calculate spring constant in the proportional region using F = ke.
Model scope: 8463 and 8464 · Foundation and Higher
What to notice: use the readout to describe how one variable changes when the other is controlled.
Worked examples
Apply calculate extension and spring constant in the worked reasoning, then try the changed-context questions below.
Scope: 8463 and 8464 · Foundation and Higher
Extension = 5.5 cm.
Scope: 8463 and 8464 · Foundation and Higher
F = 6.0 N.
Scope: 8463 and 8464 · Foundation and Higher
The limit of proportionality is about 7 N.
Exam precision
Do this: Use extension rather than total length. Check the proportional region before applying a constant k or the triangular work-area model.
Independent practice
Write an answer before opening the marking guidance. Then edit the exact phrase or step that would gain the next mark.
Scope: 8463 and 8464 · Foundation and Higher
Original PhysicsUK mark scheme: Point 1 (1 mark): k = F/e. Point 2 (1 mark): e = 0.040 m; k = 8.0/0.040. Point 3 (1 mark): k = 200 N/m. Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.
Model answer: k = F/e. e = 0.040 m; k = 8.0/0.040. k = 200 N/m.
Acceptable answers: Expected result: k = 200 N/m. Accept the unrounded value or a correctly rounded equivalent to two or more significant figures; accept equivalent units if converted correctly. No additional measurement interval is implied.
Marking notes: Method marks remain available after an arithmetic error. Carry an earlier numerical error into a later valid step; award that method point if the unit conversion and physics are correct. A fully correct answer with unit earns all marks unless working is expressly requested.
Scope: 8463 and 8464 · Foundation and Higher
Original PhysicsUK mark scheme: Point 1 (1 mark): Ee = ½ke². Point 2 (1 mark): Ee = 0.5 × 200 × 0.040². Point 3 (1 mark): Ee = 0.16 J. Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.
Model answer: Ee = ½ke². Ee = 0.5 × 200 × 0.040². Ee = 0.16 J.
Acceptable answers: Expected result: Ee = 0.16 J. Accept the unrounded value or a correctly rounded equivalent to two or more significant figures; accept equivalent units if converted correctly. No additional measurement interval is implied.
Marking notes: Method marks remain available after an arithmetic error. Carry an earlier numerical error into a later valid step; award that method point if the unit conversion and physics are correct. A fully correct answer with unit earns all marks unless working is expressly requested.
Questions pupils ask
No. Curvature shows loss of proportionality; unloading is needed to test whether the spring returns to its original length.
Useful next steps