Afully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 54 m in 3.0 min, starting and ending at rest. the elevator’s counterweight has a mass of only 950 kg, and so the elevator motor must . what average power is required of the force the motor exerts on the cab via the cable?

The average power is calculated as 735.0 W

Solution:

As per the question:

Total mass, M = 1200 kg

Counter mass of the elevator, m = 950

Distance traveled by the elevator, d = 54 m

Time taken, t = 3 min = 180 s

Now,

To calculate the average power:

First, we find the force needed for lifting the weight:

Force, F = (M - m)g =

Now, the work done by this force:

W = Fd =

Average power is given as:

the answer is b ) square root of 5