Question

An elevator can carry a maximum load of 1600 kg (elevator + passenger) is moving up with a constant speed 3 m/s. The frictional force opposing the motion is 4000 N. The minimum power delivered by the motor to the elevator in (hp) (approximately) is

Answer 1

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

please mark me as brainly