AQA A-Level Physics Paper 1, June 2025: Question 23

✅ Correct Answer

B (5.3°)

💡 Key Knowledge

• Constant Speed: Acceleration a = 0 , which means the resultant force parallel to the slope is zero ( ΣF = 0 ).
• Component of Weight: Parallel to the slope = W sin θ acting down the slope.
• Friction: Always opposes relative motion. Since the block moves up the ramp, friction acts down the ramp.

Free-Body Force Diagram Breakdown (Parallel to Ramp):
• Up the ramp: Applied push force = 25.0 N
• Down the ramp: Friction force ( 3.0 N ) + Component of weight down the ramp ( W sin θ )

Step 1: Set up the equilibrium condition along the ramp
Because the block moves at constant speed, upward forces balance downward forces:

F_push = F_friction + W sin θ

Step 2: Substitute known numerical values

25.0 = 3.0 + 240 sin θ

Step 3: Rearrange to find sin θ

240 sin θ = 25.0 - 3.0 = 22.0

sin θ = 22.0 / 240 = 0.09167

Step 4: Calculate the angle θ

θ = arcsin(0.09167) = 5.259° ≈ 5.3°

This matches option B.

🧠 Exam Technique & Distractor Analysis

• Option A (0.12°): A classic trap resulting from calculating 28 / 240 ≈ 0.12 and forgetting to take the inverse sine ( arcsin ).
• Option C (6.0°): Results from completely ignoring friction ( sin θ = 25.0 / 240 = 0.104 → θ = 6.0° ).
• Option D (6.7°): Results from adding friction to the pushing force instead of opposing it ( sin θ = (25.0 + 3.0) / 240 = 0.1167 → θ = 6.7° ).

❌ Common Errors

• Incorrect Friction Direction: Assuming friction acts in the direction of the push rather than opposing motion.
• Trig Confusion: Using cos θ instead of sin θ for the component of weight parallel to the inclined plane. Remember: parallel is always W sin θ when θ is to the horizontal.
• Calculator Mode: Having the calculator in radian mode instead of degree mode (giving 0.092 rad ).