AQA AS Level Physics Paper 2, November 2021: Question 26

1 mark · Medium difficulty · Multiple Choice

Determine the energy transferred from a mass-spring system when a mass is lowered until it hangs stationary with an extension Delta L.

Practise this question

Question

The question shows two diagrams labelled Diagram 1 and Diagram 2. Diagram 1 shows a mass m attached to an unextended vertical spring. Diagram 2 shows the same mass lowered until it hangs stationary, with the spring having an extension of Delta L. Four multiple-choice options for the energy transferred from the mass-spring system are given: A (mg Delta L)/2, B mg Delta L, C (3mg Delta L)/2, and D 2mg Delta L.
Question text

26 A mass m is added to a vertical spring that is initially unextended, as shown in Diagram 1.

The mass is then lowered until it hangs stationary on the spring, as shown in Diagram 2.

The extension of the spring is now ΔL.

Diagram 1 Diagram 2

How much energy is transferred from the mass–spring system?

[1 mark]

mg∆L

A

B mgΔL

3mg∆L

C

D 2mgΔL

Questions 27 and 28 are about three spheres X, Y and Z.

The relative mass and relative diameter of each sphere are given in the table.

X Y Z

relative mass 1 5 1

relative diameter 1 1 5

Each sphere is dropped from rest and accelerates to its terminal speed.

Mark scheme

Show the mark scheme The mark scheme table shows question 26 corresponding to correct option A, with the answer (mg Delta L)/2.

mg L

26 A

How to answer it

Energy Transfer in a Mass-Spring System

What this question tests:
This question tests your understanding of energy conservation in vertical mass-spring systems, specifically tracking gravitational potential energy lost versus elastic potential energy gained, and accounting for the energy transferred to the surroundings when a mass settles into static equilibrium.
Question 26

Energy Transferred from the Mass-Spring System

✅ Correct Answer

Option A ( mgΔL / 2 )

💡 Key Knowledge

  • Gravitational Potential Energy (GPE) Lost: As the mass falls by distance ΔL , it loses mgΔL of gravitational potential energy.
  • Elastic Potential Energy (EPE) Gained: The spring stretches by ΔL , storing 0.5 k (ΔL)² of elastic potential energy.
  • Equilibrium Condition: At the final stationary position, the upward spring force equals the weight ( kΔL = mg ).

📐 Step-by-Step Breakdown

  1. Step 1: Write out the equation for gravitational potential energy lost by the mass:
    Loss in GPE = mgΔL
  2. Step 2: Substitute k = mg / ΔL (from Hooke's Law at equilibrium) into the elastic potential energy formula:
    EPE gained = 0.5 k (ΔL)² = 0.5 (mg / ΔL) (ΔL)² = 0.5 mgΔL
  3. Step 3: Calculate the difference between energy lost and energy stored to find the energy transferred to the surroundings (thermal/dissipated energy):
    Energy transferred = Loss in GPE - EPE gained = mgΔL - 0.5 mgΔL = 0.5 mgΔL ( mgΔL / 2 ).

❌ Common Errors

Many students incorrectly choose Option B ( mgΔL ) by forgetting that energy is stored in the spring as elastic potential energy, assuming all GPE is lost to the surroundings. Others choose Option D by miscalculating work done values during initial release.

🧠 Exam Technique & Examiner Insights

Questions involving systems oscillating or settling to equilibrium frequently trap students who confuse the maximum extension during a drop ( 2ΔL ) with the final static extension ( ΔL ). Always distinguish between dynamic overshoot and final static equilibrium states.

Mark allocation: [1 mark] awarded for selecting option A.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.