Question

A lift of mass 920 kg has a capacity of 10 persons. If average mass of person is 68 kg. Friction force between lift and lift shaft is 6000 N. The minimum power of motor required to move the lift upward with constant velocity 3 m/s is [g = 10 m/s²]

(1) 66000 W 

(2) 63248 W 

(3) 48000 W 

(4) 56320 W

Answer 1

Answer is (1)

Net force on motor will be 

F_m = [920 + 68(10)]g + 6000 

= 22000 N 

So, required power for motor

P_m = vectorF_m.vectorv

= 22000 x 3 

 = 66000 watt