Question
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, ...)
Answers
drwls
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
P = [(1/2)MV^2 + M g H]/15.7 s
g = 9.8 m/s^2
H = 49.0 m
drwls
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.
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.
rory
yeah the only thing you did wrong was the final answer it should be 34.72E3 W
drwls
I did the MgH term wrong, and agree with your answer.