A car (m = 690.0 kg) accelerates uniformly from rest up an inclined road which rises uniformly, to a height, h = 49.0 m. Find the average power the engine must deliver to reach a speed of 24.9 m/s at the top of the hill in 15.7 s(NEGLECT frictional losses: air and rolling, ...)

Answer 1

Divide the energy acquired at the top of the hill (kinetic PLUS potential) buy the leapsed time (15.7 s). Energy divided by time equals power.

P = [(1/2)MV^2 + M g H]/15.7 s
g = 9.8 m/s^2
H = 49.0 m

Answer 2

Divide the energy acquired at the top of the hill (kinetic PLUS potential) by the elapsed time (15.7 s). Energy divided by time equals power.

P = [(1/2)MV^2 + M g H]/15.7 s
g = 9.8 m/s^2
H = 49.0 m
M = 690 kg
V = 24.9 m/s
P = (213,900 + 34,290)/15.7 = 18,808 W = 21 horsepower

Check my thinking and numnbers.

Answer 3

yeah the only thing you did wrong was the final answer it should be 34.72E3 W

Answer 4

I did the MgH term wrong, and agree with your answer.