Posted by Nick on .
A(n) 563 kg elevator starts from rest. It moves upward for 3.18 s with a constant acceleration until it reaches its cruising speed of 1.96 m/s. The acceleration of gravity is 9.8 m/s 2 . Find the average power delivered by the elevator motor during this period. Answer in units of kW

Physics 
drwls,
The elevator accelerates at a = 1.96/3.18 = 0.6134 m/s^2
The height increase during the acceleration period is
H = (1/2) a t^2 = 3.10 m
The energy gain during that interval is
M*g*H + (M/2)*V^2
Divide by the time interval for the average power