A car of mass 1470 kg traveling at 22 m/s is at the foot of a hill that rises 10

Question

A car of mass 1470 kg traveling at 22 m/s is at the foot of a hill that rises 100 m in 3.6 km. At the top of the hill, the speed of the car is 10 m/s. Find the average power delivered by the car's engine, neglecting any frictional losses.

Vi = 22 m/s m = 1470 Kg h = 100m D = 3600 m

Energy at the bottom Eb = (1/2)1470*(22^2) (potential Energy is zero) = 355740 J

Energy at the top Et = KE + PE = (1/2)1470(10^2) + 1470*9.8*100 = 73500+1440600 = 1514100J

Eb + Work Done = Et

355740 + W = 1514100

or Work Done W = 1514100 - 355740 = 1158360 J

time t = distance/average velocity = 3600/(22+10)/2 = 3600/16 = 225 secs

Average Power = W/t = 1158360/225 = 5148.27 Watts,

What is wrong with the above?

Explanation / Answer

First, in all questions like this, a drawing is a great help.
Draw a right-angled triangle ABC, angle ABC = 90

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