Afully loaded, slow-moving freight elevator has a cab with a total mass of 1700 kg, which is required to travel upward 45 m in 2.1 min, starting and ending at rest. the elevator's counterweight has a mass of only 840 kg, so the elevator motor must pull the cab upward. what average power is required of the force the motor exerts on the cab via the cable?

3,010 W

Explanation:

mass of elevator (Me) = 1700 kg

mass of counter weight (Mc) = 840 kg

travel distance (d) = 45 m

time (t) = 2.1 min = 126 s

What average power is required of the force the motor exerts on the cab via the cable?

from Newtons second law of motion

force exerted by the elevator motor + force exerted by the elevator counter weight = force exerted by the elevator cab weight

therefore

force exerted by the elevator motor =  force exerted by the elevator counter weight - force exerted by the elevator cab weight

force exerted by the elevator motor (Fm) = (Me x g) - (Mc x g)

force exerted by the elevator motor (Fm) =  (1700 x 9.8) - (840 x 9.8)

force exerted by the elevator motor (Fm) = 8428 N

average power exerted by the motor = Fm x speed

where speed = distance / time

average power exerted by the motor = Fm x (distance / time)

average power exerted by the motor = 8428 x (45/126) = 3,010 W

d is the answer i came up with