AQA A-Level Physics Paper 1, June 2024: Question 26

1 mark · Medium difficulty · Multiple Choice

Calculate the average power developed by a spring of stiffness k compressed by force F and returning to its original length over time t.

Practise this question

Question

Multiple choice question 26. A spring is compressed by force F, having stiffness k and length change Delta L, returning to its original length in time t. Four options are given for the average power developed: A is (k Delta L)/(2t), B is (k Delta L)/t, C is (k (Delta L)^2)/(2t), and D is (k (Delta L)^2)/t.
Question text

26 A spring is compressed by a force F. The spring has stiffness k and its length changes

by ΔL during the compression. When the force is removed the spring returns to its original

length in time t.

What is the average power developed by the spring as it returns to its original length?

[1 mark]

k∆L

A

2t

k∆L

B

t

k(∆L)2

C

2t

k(∆L)2

D

t

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer is C, with the expression (k(Delta L)^2)/(2t) classified under assessment objective AO2.

k(∆L)2

26 C AO2

2t

How to answer it

Average Power Developed by a Spring

What this question tests

This question assesses your ability to combine definitions of mechanical work, elastic potential energy stored in a spring, and average power. You must correctly recall energy equations involving spring stiffness ( k ) and extension/compression ( ΔL ), and link them to power as energy transferred per unit time ( t ).

Question 26 • Multiple Choice • 1 Mark

Full Breakdown & Examiner Insight

✅ Correct Answer: C

The correct option is C: k(ΔL)² / (2t)

Marks: 1/1 available for selecting C.

💡 Key Knowledge Required

  • Elastic Potential Energy (E): Stored energy in a compressed or stretched spring is given by E = 0.5 × k × (ΔL)² (derived from the area under a force-extension graph).
  • Power (P): Defined as rate of transfer of energy, P = Energy / time ( P = E / t ).

📐 Step-by-Step Derivation

  1. Identify total energy stored: When compressed by force F resulting in change in length ΔL , the work done stores elastic potential energy: E = 0.5 k (ΔL)² .
  2. Apply power formula: Average power is total energy transferred divided by the time taken ( t ): P = E / t .
  3. Substitute and simplify: P = [0.5 k (ΔL)²] / t , which can be rewritten as k(ΔL)² / (2t) .

🧠 Exam Technique

In multi-choice questions involving squared terms (like (ΔL)² ), eliminate options missing the square immediately (such as A and B). Then, look out for the factor of 1/2 which comes from the triangular area under a force-extension graph ( E = 0.5 × base × height ).

❌ Common Errors & Traps

  • Missing the factor of 2: Choosing option D ( k(ΔL)² / t ) by forgetting that elastic potential energy includes the 0.5 factor.
  • Forgetting the square: Choosing option A or B by treating energy simply as F × ΔL without substituting F = kΔL , missing that extension is squared in energy calculations.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.