AQA AS Level Physics Paper 2, June 2025: Question 24

1 mark · Medium difficulty · Multiple Choice

Determine the expression for the elastic strain energy stored in a stretched rod in terms of its Young modulus, cross-sectional area, length, and extension.

Practise this question

Question

Question 24 states: A rod is made of a metal of Young modulus E. When the rod is unstretched, it has a cross-sectional area A and a length L. A force is then applied to the rod that produces an extension x. The rod obeys Hooke’s law. What is the elastic strain energy stored in the rod? Four multiple-choice options are given: A is (Ax)/(2EL), B is (AEx)/(2L), C is (Ax^2)/(2EL), and D is (AEx^2)/(2L).

Mark scheme

Show the mark scheme Mark scheme for question 24 shows the correct answer is D, with the formula (AEx^2)/(2L), assessing AO2.

How to answer it

Elastic Strain Energy Stored in a Stretched Rod

📌 What this question tests

This question assesses your ability to combine foundational mechanics equations algebraically:

  • Hooke's Law & Elastic Strain Energy: Stored energy as the area under a force–extension graph ( E = ½Fx ).
  • Young Modulus definition: Linking tensile stress and tensile strain ( E = (F/A) / (ΔL/L) ).
  • Algebraic substitution & elimination: Expressing energy purely in terms of the given parameters ( E , A , L , and extension x ).

Question 24

Multiple Choice (1 Mark) — Assessment Objective 2 (AO2)

✅ Correct Answer

Correct Option: D

(AEx²) / (2L)

💡 Key Knowledge

  • Elastic Strain Energy:
    Energy = ½Fx (since the rod obeys Hooke’s law).
  • Tensile Stress:
    σ = F / A
  • Tensile Strain:
    ε = x / L
  • Young Modulus:
    E = σ / ε = (F · L) / (A · x)

📐 Step-by-Step Derivation

  1. Start with the fundamental energy equation:
    For any material obeying Hooke’s law, elastic strain energy is given by:
    Energy = ½ · F · x
  2. Express Young Modulus ( E ) in terms of the given variables:
    E = Tensile Stress / Tensile Strain = (F / A) / (x / L) = (F · L) / (A · x)
  3. Rearrange this definition to express force ( F ) in terms of E , A , L , and x :
    F = (E · A · x) / L
  4. Substitute F back into the strain energy formula:
    Energy = ½ · [ (E · A · x) / L ] · x
    Energy = (A · E · x²) / (2L)

❌ Common Errors & Traps

  • Selecting Option B ( AEx / 2L ): Forgetting that multiplying force by extension gives another factor of x . Option B has the wrong dimensions (dimensions of force, not energy).
  • Inverting E (Options A & C): Rearranging E = FL / Ax incorrectly as F = x / (EAL) or putting E in the denominator.
  • Confusing stiffness ( k ) with Young Modulus ( E ): While Energy = ½kx² , remember that k = EA / L , meaning EA / L directly substitutes for k .

🧠 Exam Technique & Dimensional Check

  • Dimensional Analysis Shortcut:
    Strain energy must be measured in Joules (N·m).
    Since E has units N m⁻², check dimensions:
    [AEx² / 2L] = (m² · (N m⁻²) · m²) / m = N·m = J ✅
    Options A and B do not yield Joules!
  • Spring Analogy:
    Recognise that a uniform rod acts like a spring where stiffness k = EA / L . Applying ½kx² gives ½(EA/L)x² = AEx² / 2L in seconds!
Examiner Insight: This 1-mark question tests pure AO2 fluency. Top-performing candidates use the spring stiffness analogy ( k = EA/L ) or dimensional analysis to arrive at D in under 30 seconds, preserving valuable time for longer calculation problems.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.