AQA AS Level Physics Paper 2, June 2025: Question 31

1 mark · Medium difficulty · Multiple Choice

Determine the energy transferred to an aircraft engine per second given constant velocity, resistive force, and engine efficiency.

Practise this question

Question

Multiple choice question asking: 'An aircraft flies horizontally with a constant velocity v. The aircraft is subject to a constant resistive force F. The aircraft engine has an efficiency of 10%. How much energy must be transferred to the engine in one second?' Options are: A: 10Fv, B: 11Fv, C: 10Fv squared, D: 11Fv squared.

Mark scheme

Show the mark scheme Mark scheme table row for question 31 showing the correct answer as option A (10Fv) with assessment objective AO2.

How to answer it

Aircraft Energy Input & Efficiency

📋 What this question tests

This question assesses your ability to apply the mechanical power equation ( P = Fv ), relate power to rate of energy transfer, and correctly manipulate the formula for percentage efficiency to deduce total input energy per second.

Question 31

Multiple Choice Analysis

Topic: Work, Energy, Power & Efficiency (AO2 Application)

✅ Correct Answer

A: 10Fv

Award: 1 Mark (AO2)

💡 Key Knowledge

  • Newton's First Law: At constant velocity, net horizontal force = 0. Therefore, engine forward thrust = resistive force = F .
  • Mechanical Power: P = Fv (useful output power delivered by the engine).
  • Power & Energy: Power is work done per unit time ( P = E / t ), so energy transferred in 1 second is numerically equal to power ( E = P × 1 ).
  • Efficiency Definition:
    Efficiency = (Useful Power Output / Total Power Input)

📐 Step-by-Step Derivation

  1. Identify the forward thrust: Because the aircraft travels at constant horizontal velocity v , it is in equilibrium. The forward thrust provided by the engines must equal the resistive force:
    Thrust = F
  2. Calculate useful power output:
    Puseful = Force × Velocity = Fv
  3. Apply the efficiency formula:
    Efficiency = 10% = 0.10
    Efficiency = Puseful / Pinput
    0.10 = Fv / Pinput
  4. Rearrange for input power:
    Pinput = Fv / 0.10 = 10Fv
  5. Find energy input per second ( t = 1 s ):
    Einput = Pinput × t = 10Fv × 1 = 10Fv

❌ Common Errors & Distractor Traps

  • Confusing Power with Kinetic Energy (Options C & D): Students seeing v² often falsely conflate P = Fv with kinetic energy formulas ( ½mv² ) or quadratic aerodynamic drag relationships. The resistive force is explicitly given as a constant F , so power depends strictly linearly on velocity: Fv .
  • Inverting the Efficiency Relationship (Options B & D): Thinking that a 10% loss means you add 10% to the useful energy (i.e. 1.1 × Fv or 11/10 ), giving 11Fv . Remember: if only 10% of total input reaches output, total input must be 1 / 0.10 = 10 times larger, not 1.1 times!
  • Overlooking "in one second": Misunderstanding that "energy in one second" is directly the definition of power in watts (Joules per second).

🧠 Exam Technique

  • Dimensional Checking: Notice dimensions of force × velocity = N × m s⁻¹ = J s⁻¹ = W . Multiplying by 1 s gives Joules ( J ). Any option with v² has dimensions of J m s⁻¹ , which is physically impossible for energy! Eliminating C and D instantly reduces the question to a 50/50 choice.
  • Sanity Check Values: If useful output is 100 J and efficiency is 10%, you need 1000 J input. 1000 / 100 = 10 , confirming the factor is 10, not 11.

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.