Question

An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.20 m/s. The minimum horsepower of the motor is: (assume g = 9.8 m/sec²)
1. 2.2 HP
2. 2 HP
3. 1.5 HP
4. 1.3 HP

Answer 1

Correct Answer - Option 4 : 1.3 HP

Concept:

As the elevator is going up with a uniform velocity V, the total work done on it is zero in any time interval.

The work done by the motor is, therefore, equal to the work done by the force of gravity in that interval.

The rate of doing work, i.e., the power delivered is ∴ P = Force × Velocity = mg × V

Calculation:

Given:

m = 500 kg, V = 0.20 m/s, g = 9.81 m/s²

P = mg × V = 500 × 9.81 × 0.20 = 981 W

As we Know 1 HP = 746 W

\(\therefore {\rm{P}} = 981 \times \frac{1}{{746}} = 1.3{\rm{\;HP}}\)

The minimum horsepower of the motor is 1.3 HP.