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Specification: AQA A-level Physics (7408) 3.4.2.1  |  Paper: 1 (and 2 as assumed knowledge)  |  Data and formulae booklet: given in the exam

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The equation

force = stiffness constant × extension    F = kΔL

Quantity	Symbol	Unit
force	F	newtons (N)
stiffness constant (spring constant)	k	newtons per metre (N m⁻¹)
extension (or compression)	ΔL	metres (m)

Rearranged: k = F ÷ ΔL  |  ΔL = F ÷ k  |  ΔL is the change in length (new length − original length), always in metres.

Use FIFA for every answer

• F – Formula: write the equation as it appears on the sheet.
• I – Insert: put in the numbers (convert to standard units first).
• F – Fix: rearrange to make the unknown the subject.
• A – Answer: calculate and always give the unit.

Core: Q1–4.   Exam standard: Q5–8.   Stretch: Q9–10.

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Questions

Q1 (Core) A spring has a stiffness constant of 40 N m⁻¹. Calculate the force needed to extend it by 0.15 m.

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F: F = kΔL
I: F = 40 × 0.15
F: F is already the subject
A: 6.0 N

Q2 (Core) A force of 12 N extends a spring by 0.080 m. Calculate the stiffness constant of the spring.

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F: F = kΔL
I: 12 = k × 0.080
F: k = 12 ÷ 0.080
A: 150 N m⁻¹

Q3 (Core) The length of a spring increases from 120 mm to 156 mm when a 7.2 N force is applied. Calculate its stiffness constant.

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Convert first: ΔL = 156 − 120 = 36 mm = 0.036 m
F: F = kΔL
I: 7.2 = k × 0.036
F: k = 7.2 ÷ 0.036
A: 200 N m⁻¹
Watch out: using the new length (0.156 m) instead of the extension is one of the most common errors.

Q4 (Core) A car suspension spring has a stiffness constant of 35 kN m⁻¹. Calculate how far it is compressed, in mm, by a force of 2.8 kN.

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Convert first: 35 kN m⁻¹ = 35 000 N m⁻¹; 2.8 kN = 2800 N
F: F = kΔL (Hooke’s law applies to compression as well as extension)
I: 2800 = 35 000 × ΔL
F: ΔL = 2800 ÷ 35 000
A: 0.080 m = 80 mm

Q5 (Exam) A force–extension graph for a spring (force on the y-axis) is a straight line through the origin. One point on the line is at an extension of 48 mm and a force of 3.6 N. Calculate the stiffness constant, then the extension produced by a force of 2.1 N, in mm.

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Convert first: 48 mm = 0.048 m
Step 1 – k from the gradient: k = ΔF ÷ ΔL = 3.6 ÷ 0.048 = 75 N m⁻¹
Step 2 – extension
F: F = kΔL
I: 2.1 = 75 × ΔL
F: ΔL = 2.1 ÷ 75
A: 0.028 m = 28 mm
Watch out: convert the extension to metres before finding the gradient, or k comes out 1000 times too small.

Q6 (Exam) A spring with a stiffness constant of 60 N m⁻¹ reaches its limit of proportionality at a force of 9.0 N. Calculate the extension at the limit of proportionality. A student uses F = kΔL to predict an extension of 0.25 m for a 15 N force. Explain why this prediction is unreliable.

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F: F = kΔL
I: 9.0 = 60 × ΔL
F: ΔL = 9.0 ÷ 60
A: 0.15 m
15 N is beyond the limit of proportionality, so force is no longer directly proportional to extension and F = kΔL does not apply. The spring may also pass its elastic limit and be permanently deformed.

Q7 (Exam) A student plots extension (y-axis) against force (x-axis) for a spring. The straight line passes through the origin and the point (20 N, 80 mm). Calculate the stiffness constant, then the force needed for an extension of 18 mm.

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Convert first: 80 mm = 0.080 m; 18 mm = 0.018 m
Step 1 – gradient: ΔL ÷ F = 0.080 ÷ 20 = 4.0 × 10⁻³ m N⁻¹, so k = 1 ÷ gradient = 250 N m⁻¹
Step 2 – force
F: F = kΔL
I: F = 250 × 0.018
F: F is already the subject
A: 4.5 N
Watch out: with extension on the y-axis the gradient is 1 ÷ k, not k.

Q8 (Exam) A spring of natural length 20.0 cm and stiffness constant 45 N m⁻¹ hangs vertically. A 0.60 kg mass is hung from it and comes to rest. Calculate the new length of the spring in m, to 2 s.f. (Assume the limit of proportionality is not reached.)

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Step 1 – force: F = mg = 0.60 × 9.81 = 5.886 N
Step 2 – extension
F: F = kΔL
I: 5.886 = 45 × ΔL
F: ΔL = 5.886 ÷ 45
A: ΔL = 0.1308 m, so new length = 0.200 + 0.1308 = 0.3308 m = 0.33 m (2 s.f.)
Watch out: the question asks for the length, not the extension. Add the extension to the natural length.

Q9 (Stretch) Two springs, k₁ = 30 N m⁻¹ and k₂ = 60 N m⁻¹, are joined end to end (in series) and a 4.0 N load is hung from the bottom. Calculate the combined stiffness constant and the total extension, then check your answer by finding the extension of each spring.

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Step 1 – combined stiffness
F: 1/k = 1/k₁ + 1/k₂
I: 1/k = 1/30 + 1/60 = 3/60
F: k = 60 ÷ 3
A: 20 N m⁻¹
Step 2 – total extension: ΔL = F ÷ k = 4.0 ÷ 20 = 0.20 m
Check: each spring carries the full 4.0 N, so ΔL₁ = 4.0 ÷ 30 = 0.133… m and ΔL₂ = 4.0 ÷ 60 = 0.0666… m; total = 0.20 m
Watch out: in series each spring carries the whole load, so the combination is less stiff than either spring on its own.

Q10 (Stretch) Two identical springs, each of stiffness constant 80 N m⁻¹, hang side by side (in parallel) and are joined at the bottom to a third identical spring in series. A 0.50 kg mass hangs from the bottom spring. Calculate the total extension of the arrangement, in mm to 2 s.f.

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Step 1 – parallel pair: k = k₁ + k₂ = 80 + 80 = 160 N m⁻¹
Step 2 – pair in series with the third spring: 1/k = 1/160 + 1/80 = 3/160, so k = 53.33… N m⁻¹
Step 3 – load: F = mg = 0.50 × 9.81 = 4.905 N
Step 4 – extension
F: F = kΔL
I: 4.905 = 53.33… × ΔL
F: ΔL = 4.905 ÷ 53.33…
A: 0.09196… m = 92 mm (2 s.f.)
Watch out: in parallel the stiffness constants add; in series the reciprocals add. Mixing these up is a common error.

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