Burning of fuel transfers 5 * 10⁵ W of power into the engine of a 1500-kg vehicle;suppose the maximum speed the vehicle can achieve 5 seconds after starting from rest is 32.6 m/s.Determine the efficiency of the engine.

Question

Quizplus · Accepted Answer

Efficiency (η) can be calculated using the formula: η = (Work output / Energy input) x 100% The work output is equal to the kinetic energy gained by the vehicle: Work output = (1/2)mv^2 = (1/2)(1500 kg)(32.6 m/s)^2 = 795450 J The energy input is equal to the power delivered to the engine multiplied by the time it operates: Energy input = (Power)(Time) = (5 F1F1F1S1F1F1F10 10⁵ W)(5 s) = 2.5x10^8 J Therefore, the efficiency is: η = (795450 J / 2.5x10^8 J) x 100% = 0.31818... x 100% = 32% (rounded to 2 significant figures) The closest to this value is choice C, 32%.

Burning of Fuel Transfers 5 * 105 W of Power

Burning of Fuel Transfers 5 * 10⁵ W of Power