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16 January, 06:07

A 6861 kg car traveling at 48 m/s is at the foot of a hill that rises 142 m in 2.3 km. At the top of the hill, the speed of the car is 10 m/s. The acceleration of gravity is 9.81 m/s 2. Assuming constant acceleration, find the average power delivered by the car's engine, neglecting any internal frictional losses.

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  1. 16 January, 09:04
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    Answer: P = 25050.8w

    Explanation:

    total energy at top = K. E + P. E

    = (1/2) (6861) (100) + 6861 (9.81) (142)

    total energy at bottom

    = (1/2) (6861) (48) ^2

    work done = energy at top - energy at bottom

    average velocity = (48+10) / 2

    time = 2300/average velocity

    power = work done/time

    plus potential) at the base and the top; is the energy input from the engine

    the ascent time is the average speed, (top + bottom) / 2; divided by the 2.3 km distance

    energy / time equals power
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