fully loaded, slow-moving 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?

Question

fully loaded, slow-moving 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

bobpursley · Answer

The lifting force is (1250-1005)g or force=245g power=work/time=245*58/(3.4*60)

beez · Answer

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