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Homework Statement
A 523 kg elevator starts from rest. It moves upward for 3.79s with a constant acceleration until it reaches its cruising speed of 1.78 m/s.
The acceleration of gravity is 9.8 m/s^2.
Find the average power delivered by the elevator motor during the period of this acceleration. Answer in units of kW.
Homework Equations
w = m(a+g)
p = w/t
The Attempt at a Solution
m = 523
a = 1.78
t = 3.79
w=m(a+g)=523(1.78+9.8)=6056.34
p=w/t=6056.34/3.79=1597.98W
1597.98W becomes 1.59798kW
I was incredibly confident in this answer, but it is wrong. Where did I make my silly error?