A car with mass of 1000.0 kg accelerates from 0 m/s to 40.0 m/s in 10.0 s. Ignor

Question

A car with mass of 1000.0 kg accelerates from 0 m/s to 40.0 m/s in 10.0 s. Ignore air resistance. The engine has a 22% efficiency, which means that 22% of the thermal energy released by the burning gasoline is converted into mechanical energy. (a) What is the average mechanical power output of the engine? (b) What volume of gasoline is consumed? Assume that the burning of 1.0 L of gasoline releases 47 MJ of thermal energy.

Explanation / Answer

(a)Let the acceleration is a m/s^2 and the distance covered in 10 sec is s meter by the car, By v = u + at =>40 = 0 + a x 13 =>a = 40/13 = 3.0769 m/s^2 By s = ut + 1/2at^2 =>s = 0 + 1/2 x 3.0769 x 13 x 13 = 259.99m Thus Power(output) = W/t = (F x s)/t = (m x a x s)/t = (1000 x 3.0769 x 200)/13 = 47336.923 Watt (b)Thus power(input) =[Power(output)/Efficiency] x 100% =>power(input) = [47336.923/22] x 100 = 215167.83 Watt Thus Energy(input) by gasoline = P(input) x time = 215167.83 x 10 = 21.5167MJ Thus the gasoline consumed = 21.5167/46 = 0.46775 L

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