Question

A lift weighing \( 250 \, \text{kg} \) is to be lifted up at a constant velocity of \( 0.20 \, \text{m/s} \). What would be the minimum horsepower of the motor to be used?

Hint:

Horsepower is a practical unit for measuring power, commonly used for motors and engines.

Answer 1

Step 1: Calculate power required to lift the lift
The force required to lift the lift is equal to its weight: \[ F = mg = 250 \times 9.8 = 2450 \, \text{N} \] The power required is given by: \[ P = Fv = 2450 \times 0.20 = 490 \, \text{W} \] Step 2: Convert to horsepower
Since \( 1 \, \text{hp} = 746 \, \text{W} \), the required horsepower is: \[ \text{Power in hp} = \frac{490}{746} \approx 0.66 \, \text{hp} \]