Question

an elevator weighing 500 kg is to be lifted up at a constant velocity of 0.4m/s . what should be the minimum horse power of the motor to be used?

Answer 1

given - 
mass of lift = 500 kg
g = 10 m/s2
velocity (v) = 0.4 m/s

displacement D in 1 sec = 0.4m


to find - HP of motor used


solution - 
assuming that the motor is an ideal one-
work done = force x D
= m x g x D
= 500 x 10 x 0.4 = 2000 Nm.

time taken = 1 sec


Power = work done / time
= 2000 / 1 = 2000 W = 2kW = 2.68 HP  [ 1 watt = 0.00134 HP]

hence required hp of motor = 2.68 HP (ANS)

Answer 2

Answer:

ExplanationGiven

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