Question

An elevator weighing 10 tonne is to be lifted up at
a constant velocity of 0.6 m/s. What should be the
minimum power of the motor to be used?
(g = 10 m/s2)
(1) 20 kW
(2) 10 kW
(3) 60 kW
(4) 120 kW​

Answer 1

Answer:

Given:

⇒ Mass = 10 ton = 9071 kg

⇒ g = 9.8 m/s² (Constant)

⇒ Velocity = 0.6 m/s

So, as we know,

⇒ Force = mg

⇒ Force = 9071 × 9.8

⇒ Force = 88895.8 N

Now, Power = Force × Velocity

⇒ Power = 88895.8 × 0.6

⇒ Power = 53337.48 W.