Question

An elevator weighting 500kg is to be lifted up at a constant velocity of 0.20 m/s.what would be the minimum power of the motor to be used​

Answer 1

Answer:

Given

Force = m g = 500 kg x 9.8 m/s²=4900 N

Velocity = 0.4 m/s

Now, power=Force x velocity

=4900 x 0.4=1960 W

Now 1 W = 0.00134 hp

So, 1960 W=1960 x 0.00134=2.6 hp