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

Question

Quizplus · Accepted Answer

The efficiency of an engine is defined as the ratio of the work output to the heat input. In this case, the work output is the potential energy gained by the truck, which is mgh = (2700 kg)(9.8 m/s^2)(3.0 m) = 79,860 J. The heat input is 4.18 × 10^5 J. Therefore, the efficiency is 79,860 J / 4.18 × 10^5 J = 0.19.

An Engine Is Used to Lift a 2700-Kg Truck to a Height