Burning fuel physics problem.?
The burning of fuel transfers 3. 105 W of power into the engine of a 2300-kg vehicle. If the engine's efficiency is 23%, determine the maximum speed the vehicle can achieve 4 s after starting from rest.
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So you have the rate of energy being transferred and that the efficiency is 23%. So if you take the 3x10^5 and multiply by 0.23, you will have the power of the engine in watts.
If you multiply that by 4s, you have the work done, which is equal to the kinetic energy of the vehicle, 1/2*m*V^2
Solve for V.
0.23*3.105 = 0.714 watts transferred
1watt - 1 newton,meter/second
therefore we have 0.714 newton meters/sec
or 0.2856 newton meters in 4 sec
K.E. of the vehicle = 1/2 m V^2
equate the 2 for the answer.