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
[tex]F=mg\\F=(3.0*10^{3}kg )(9.8m/s^{2} )\\F=29400N[/tex]
The velocity of elevator is given as
[tex]v=d/t\\v=(210m)/23s\\v=9.13m/s[/tex]
Since the net force has same direction of motion so α=0°
So
[tex]P=F.VCos\alpha \\P=(29400N).(\frac{210m}{23s} ).Cos(0)\\P=268kW[/tex]
