An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.25 m/s. What would be the minimum horsepower of the motor to be used? G = 9.8 m/s² and there is no frictional loss.

Concept:

• Power is the rate at which work is done upon an object.
• Horsepower (hp) is the measurement of power.
  •  1hp = 746 watt
• Si unit of power is the watt.

Calculation:

Given,

Elevator weigh or mass (m) = 500kg

Velocity = 0.25m\s

G = 9.8 m\s²

We know that,

Power = Force × velocity

power = mg× v

power = 500× 9.8× 0.25

power = 1225 watt

we also know that 1 hp = 746 watt

Therefore, 1 watt = 1\746

So, 1225 watt = 1225 \ 746 = 1.64 hp.

Thus, the minimum horsepower of the motor used is 1.64hp.