10 AQA GCSE Physics questions on elastic potential energy (EPE = 1/2 ke^2) that get harder as you go, from Foundation to Higher, each with a full worked solution.
Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Equation sheet: given in the exam
📌 KEY DEFINITIONS – learn them word for word
Elastic potential energy store: the energy stored in an object that has been stretched or squashed, such as a spring.
Limit of proportionality: the point beyond which the extension of a spring is no longer directly proportional to the force applied. Ee = ½ke² only works up to this point.
The equation
elastic potential energy = 0.5 × spring constant × (extension)² Eₑ = ½ k e²
| Quantity | Symbol | Unit |
|---|---|---|
| elastic potential energy | Eₑ | joules (J) |
| spring constant | k | newtons per metre (N/m) |
| extension | e | metres (m) |
Rearranged: k = 2Eₑ ÷ e² | e = √(2Eₑ ÷ k) | Square the extension before multiplying.
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.
Foundation: aim for Q1–8. Higher: aim for Q4–10. Most questions need rearranging.
Questions
Q1 (F) A spring with a spring constant of 200 N/m is stretched by 0.10 m. Calculate the elastic potential energy stored.
Show solution
F: Eₑ = ½ k e²
I: Eₑ = 0.5 × 200 × 0.10²
F: Eₑ is already the subject
A: 0.5 × 200 × 0.01 = 1 J
Q2 (F) A catapult elastic stretched by 0.20 m stores 4.0 J. Calculate its spring constant.
Show solution
F: Eₑ = ½ k e²
I: 4.0 = 0.5 × k × 0.20², so 4.0 = 0.02 × k
F: k = 4.0 ÷ 0.02
A: 200 N/m
Q3 (F) A trampoline spring has a spring constant of 800 N/m and stores 16 J. Calculate its extension.
Show solution
F: Eₑ = ½ k e²
I: 16 = 0.5 × 800 × e², so 16 = 400 × e²
F: e² = 0.04, so e = √0.04
A: 0.2 m
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Q4 (F) A toy launcher spring (k = 50 N/m) stores 0.09 J. How far is it compressed, in cm?
Show solution
F: Eₑ = ½ k e²
I: 0.09 = 0.5 × 50 × e², so 0.09 = 25 × e²
F: e² = 0.0036, so e = 0.06 m
A: 6 cm
Q5 (F) A car suspension spring (k = 40 000 N/m) stores 50 J when compressed. How far is it compressed, in cm?
Show solution
F: Eₑ = ½ k e²
I: 50 = 0.5 × 40 000 × e², so 50 = 20 000 × e²
F: e² = 0.0025, so e = 0.05 m
A: 5 cm
Q6 (F/H) An archer’s bow acts like a spring with k = 1500 N/m. When drawn back it stores 120 J. How far is the string pulled back?
Show solution
F: Eₑ = ½ k e²
I: 120 = 0.5 × 1500 × e², so 120 = 750 × e²
F: e² = 0.16, so e = √0.16
A: 0.40 m
Q7 (F/H) At the bottom of a jump, a bungee cord is stretched 20 m and stores 12 kJ. Calculate its spring constant.
Show solution
Convert first: 12 kJ = 12 000 J
F: Eₑ = ½ k e²
I: 12 000 = 0.5 × k × 20², so 12 000 = 200 × k
F: k = 12 000 ÷ 200
A: 60 N/m
Q8 (F/H) A train buffer spring has k = 2.0 × 106 N/m and stores 1.0 × 104 J. Calculate its compression.
Show solution
F: Eₑ = ½ k e²
I: 1.0 × 104 = 0.5 × 2.0 × 106 × e², so 1.0 × 104 = 1.0 × 106 × e²
F: e² = 0.01, so e = √0.01
A: 0.10 m (1.0 × 10−1 m)
Q9 (H) A catapult (k = 400 N/m) is stretched 0.15 m and fires a 20 g stone. Assuming all the elastic energy becomes kinetic energy, calculate the stone’s speed to 2 significant figures.
Show solution
Step 1 – energy stored: Eₑ = ½ k e² = 0.5 × 400 × 0.15² = 4.5 J
Step 2 – speed
F: Eₖ = ½ m v²
I: 4.5 = 0.5 × 0.020 × v², so 4.5 = 0.010 × v²
F: v² = 450, so v = √450 = 21.2… m/s
A: 21 m/s (2 s.f.)
Q10 (H) A force of 30 N stretches a gym spring by 0.12 m. Calculate the elastic potential energy stored.
Show solution
Step 1 – spring constant
F: F = k e
I: 30 = k × 0.12
F: k = 30 ÷ 0.12 = 250 N/m
Step 2 – energy
F: Eₑ = ½ k e²
I: Eₑ = 0.5 × 250 × 0.12²
A: 1.8 J
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