Answer:
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