AQA AS Level Physics Paper 2, June 2025: Question 34

1 mark · Medium difficulty · Multiple Choice

Calculate the minimum power output of a car's engine moving at constant speed up a slope, given its weight, speed, and horizontal resistive forces.

Practise this question

Question

Multiple-choice question showing a diagram of a car moving from a horizontal surface onto an inclined slope of 10 degrees to the horizontal. The question states that the car moves at a constant speed of 15 m s⁻¹ with an engine thrust of 3800 N on the horizontal road, the car has a weight of 12 000 N, and resistive forces remain constant. Four choices for the minimum power output up the hill are given: A 31 kW, B 57 kW, C 88 kW, D 230 kW.

Mark scheme

Show the mark scheme Mark scheme row for question 34 showing correct answer C (88 kW) mapped to assessment objective AO2.

How to answer it

Car Engine Power on an Inclined Slope

📋 What this question tests

This question assesses your ability to apply Newton's first law of motion to find unknown resistive forces, resolve a weight vector parallel to an inclined plane, determine the total required driving thrust for constant velocity, and calculate mechanical power using P = Fv.

Question 34 Analysis

Multiple Choice [1 Mark] • Assessment Objective: AO2 (Application of Knowledge)

✅ Correct Answer

C — 88 kW

The engine must overcome both the constant resistive force (3800 N) and the component of the car's weight acting down the slope (2084 N), giving a total thrust of ~5884 N at a speed of 15 m s⁻¹.

💡 Key Knowledge

  • Newton's First Law: At constant speed along a straight line, net force = 0. Forward thrust = total opposing force.
  • Resolving Weight on a Slope: Component parallel to slope acting downwards = W sin(θ) .
  • Power Formula: P = F × v , where F is the forward thrust and v is the constant speed.

📐 Step-by-Step Calculation

Step 1: Determine the resistive forces from horizontal travel

On a horizontal road at constant speed (15 m s⁻¹):

Resultant force = 0 ⇒ Fresistive = Thrust = 3800 N

Since the question states resistive forces remain constant, Fresistive = 3800 N on the hill as well.

Step 2: Resolve the weight component acting parallel down the slope

The car climbs a hill inclined at 10° to the horizontal:

Fslope = W × sin(θ) = 12 000 × sin(10°)

Fslope = 12 000 × 0.17365 = 2083.8 N

Step 3: Calculate total forward thrust required to maintain constant speed

To continue moving at a constant speed of 15 m s⁻¹ up the hill, the engine thrust must balance both opposing forces:

Ftotal = Fresistive + Fslope = 3800 + 2083.8 = 5883.8 N

Step 4: Calculate power output

Using the mechanical power formula:

P = Ftotal × v = 5883.8 N × 15 m s⁻¹ = 88 257 W ≈ 88 kW

🧠 Exam Technique & Traps Decoded

Each incorrect distractor corresponds to a classic student shortcut or calculation trap:

  • 31 kW (Option A): Calculated power against weight only: (12 000 × sin 10°) × 15 ≈ 31 kW . Forgot to include the 3800 N resistive forces!
  • 57 kW (Option B): Horizontal power only: 3800 × 15 = 57 kW . Completely ignored the slope.
  • 230 kW (Option D): Added the full un-resolved weight to thrust: (3800 + 12 000) × 15 ≈ 237 kW . Failed to resolve parallel to the slope.

❌ Common Errors to Avoid

  • Using cos(θ) instead of sin(θ): Remember that the slope component is opposite the angle θ in the force triangle — always use W sin(θ) along the incline.
  • Mass vs Weight confusion: The question provides weight directly as 12 000 N. Do not multiply by g (9.81 m s⁻²) again!
  • Forgetting units: The calculated value is ~88 257 W; divide by 1000 to convert to kW.
Mark Scheme Breakdown: Award [1 mark] for selecting option C (88 kW).

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.