A tractor pulls a trailer along a level road at a steady 4.0 m s−1 with a horizontal force of 1500 N. Its engine supplies 10 kW. Find the useful power and the efficiency.
- Useful power = force × velocity = 1500 × 4.0 = 6000 W = 6.0 kW.
- Efficiency = useful output ÷ total input = 6.0 ÷ 10 = 0.60.
- As a percentage, 0.60 × 100% = 60%.
Answer: Useful power 6.0 kW; efficiency 60%.