Power against gravitational force on incline

Steps:

• Total mass m = 80,000 kg (train) + 200 × 70 kg (passengers) = 94,000 kg.
• Gravitational force component parallel to slope F = m g sin 30° = 94,000 × 10 × 0.5 = 470,000 N (using g = 10 m/s²).
• Power to raise load = F × v = 470,000 × 5 = 2,350,000 W (using v = 5 m/s along slope).
• Engine power = raising power / 0.4 = 2,350,000 / 0.4 = 5,875,000 W ≈ 6 MW.

Why C is correct:

• It matches the total engine power calculated from the 40% fraction of input power converted to work against m g sin θ v, per the work-energy principle.

Why the others are wrong:

• A: Underestimates by assuming no efficiency loss or lower mass/speed.
• B: Matches raising power alone, ignoring the 40% means total is 2.5 times larger.
• D: Overestimates by doubling mass or speed beyond given values.

Final answer: C