Work, Energy & Power — A/L Physics
Aligned with the NIE syllabus
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1 question with a worked solution
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An electric locomotive pulls a train of total mass 2.0 × 10⁵ kg up an incline of 1 in 100 (sin θ = 0.01) at a constant steady speed of 20 m s⁻¹. Total resistive friction forces amount to 1.5 × 10⁴ N. Taking g = 10 m s⁻², what output power must the locomotive develop?
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Correct answer: D) 700 kW
Tractive force F must balance the downhill component of gravity plus frictional resistance: F = mg sin θ + Ffriction = (2.0 × 10⁵ kg × 10 m s⁻² × 0.01) + 1.5 × 10⁴ N = 2.0 × 10⁴ N + 1.5 × 10⁴ N = 3.5 × 10⁴ N. Output Power P = F × v = 3.5 × 10⁴ N × 20 m s⁻¹ = 7.0 × 10⁵ W = 700 kW.
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Step 1: Downward gravity component = mg sin θ = 2.0 × 10⁵ × 10 × 0.01 = 20,000 N. Step 2: Total required force F = 20,000 N + 15,000 N = 35,000 N. Step 3: Power P = F × v = 35,000 N × 20 m s⁻¹ = 700,000 W = 700 kW. Answer: 700 kW.Modelled on: the A/L Physics examination pattern
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Where does Work, Energy & Power fit in the syllabus?
Unit 2.4 · Year 1 · Term 1
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