Question

The loaded cab of an elevator has a mass of 3.0 × 103 kg and moves 210 m up the shaft in 23 s at constant speed. At what average rate does the force from the cable do work on the cab?

Answer 1

Step-by-step explanation:

Given data

The elevator mass=3.0×10³ kg

Time t=23 s

elevator lift d=210 m

The power is the average rate of work done:

So

P=F.V Cosα

Where F is force

V is velocity

α is angle between Force and velocity

Apply the Newton Law to find the force on elevator

`F=mg\\F=(3.0*10^(3)kg )(9.8m/s^(2) )\\F=29400N`

The velocity of elevator is given as

`v=d/t\\v=(210m)/23s\\v=9.13m/s`

Since the net force has same direction of motion so α=0°

So

`P=F.VCos\alpha \\P=(29400N).((210m)/(23s) ).Cos(0)\\P=268kW`