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

2 answers

The lifting force is (1250-1005)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