Question

An engine is used to lift a 2700-kg truck to a height of 3.0m at constant speed. In the lifting . process, the engine received 4.18 x 105 J of heat from the fuel burned in its interior. What is the efficiency of the engine?

Answer 1

The efficiency of the engine is given by $k=\frac{mgh}{q}=\frac{2700 \cdot 9.8 \cdot 3}{4.18 \cdot 10^5}\approx 0.19$