CATAGORY: Mechanics QUESTION: I seem to get different answers from this for this question. A small car travelling at a constant speed of 20ms-1 on a horizontal road is subjected to air resistance of 570N and road friction of 150N. a) What power providded by the engine of the car is used to keep it in motion at this speed? b) If the effieciency of the car’s engine is 25%, at what rate is the chemical energy in the fuel being transformed? c) If 1 Litre of fuel provides 3.3 ´ 107 J of chemical energy when it is burned, how much fuel is used by the car if it travels at this speed for 100km?

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----> -----> (1) Power = Force * velocity Because the force is in the opposite direction as the velocity the Power is negative meaning a power drain. The total frictional force is 570N + 150N = 720N so at 20m/s the Power loss is (2) Power = -(720N) 20m/s = -14400 J/s The car engine must then generate +14400 J/s to compensate for the loss. Since the car engine is only 25% efficient the then four times as much (1/.25) chemical energy must be transformed so the rate of chemical energy conversion is 4(14400 J/s) = 57600 J/s. At 20m/s 100km is traveled in (3) t = (100km x 1000m/km)/20m/s = 5000 s ~ 1.38 hours During this time the amount of chemical energy used up is (4) E = 57600 J/s x 5000 s = 2.88 x 10^8 J If 1 litre of fuel provides 3.3 x 10^7 J then the number of litres used up is (5) L = 2.88 x 10^8 J/ (3.3 x 10^7 J/l) = 8.73 lietres ********************* * * * (a) 14400 J/s * * * * (b) 57600 J/s * * * * (c) 8.73 lietres * * * *********************