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.

Question

Drew

Physics

23 August, 07:13

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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Answers (1)

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Madden Irwin · Answer 1

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