College Physics Power Problem?
The loaded cab of an elevator has a mass of 2.2 x 103 kg and moves 201 m up the shaft in 22 s at constant speed. At what average rate does the force from the cable do work on the cab?
More Questions From This User See All
Answers & Comments
Verified answer
F = force in cable = mg since there is no acceleration.
W = Fd
rate at which work is done = work / time
divide both sides by time
W/t = Fd/t
= mg(d/t)
= (2.2 x 10^3 kg) (9.81 m/s^2) (201 m)/ (22s)
= 197 181 J/s