Question

. An elevator can carry a maximum load of 1800 kg
(elevator + passengers) is moving up with a constant
speed of 2 m/s^2.The frictional force opposing
the motion is 4000 N. What is minimum power
delivered by the motor to the elevator?​

Answer 1

Answer:

Given that,

Mass of elevator + passengers, M=1800Kg

Speed, v=2m/s

Friction force, f=400N

Power delivered, P=force×velocity.........(1)

Force acting downward,

F=mg+f

F=1800×10+4000

F=22000N

Put the value of F in equation (1)

P=22000×2

P=44000Watts