Question

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1400 kg, which is required to travel upward 37 m in 3.6 min, starting and ending at rest. The elevator's counterweight has a mass of only 930 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable?

Answer 1

789.8 W

Step-by-step explanation:

mass of the cab = 1400 kg, the counter weight of the elevator = 930 kg

weight of the cab = 1400 × 9.81 where weight = mg and m is mass and g is acceleration due to gravity.

weight of the cab = 13734 N

counter weight of the elevator = 930 × 9.81 = 9123.3 N

the exerted force of the elevator = weight of the cab - counter weight of the elevator = 13734 - 9123.3 = 4610.7 N

Average power by the motor P = F × v = F × distance / time

where v is speed in m/s, and time is in seconds

P = 4610.7 × 37 / ( 3.6 × 60) = 789.80 W

where (3.6 × 60 ) is the time in seconds