Physics
posted by Corey .
fully loaded, slowmoving freight elevator has a cab with a total mass of 1250 kg, which is required to travel upward 58 m in 3.4 min, starting and ending at rest. The elevator's counterweight has a mass of only 1005 kg, and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?
________W

The lifting force is (12501005)g
or force=245g
power=work/time=245*58/(3.4*60) 
force of gravity on elevator is
1250 kg * 9.8= 12250 N
Force of counter weight is
1005 kg * 9.8= 9849 N
12250 N  9849 N = 2761 N Total
P = (F*d)/T
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