Question

A fully 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 help. What average power is required of the force the motor exerts on the cab via the cable?

Answer 1

Answer:

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 = $(1200 - 950)\times 9.8 = 2450 N$

Now, the work done by this force:

W = Fd = $2450\times 54 = 132300\ J = 132.3\ kJ$

Average power is given as:

$P_{avg} = \frac{W}{t} = \frac{132300}{180} = 735.0\ W$