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: Determine power for lifting the lift.
The force needed to lift the lift equals its weight:
\[F = mg = 250 \times 9.8 = 2450 \, \text{N}\]
Power is calculated as:
\[P = Fv = 2450 \times 0.20 = 490 \, \text{W}\]

Step 2: Convert to horsepower.
Using the conversion \( 1 \, \text{hp} = 746 \, \text{W} \), the required horsepower is:
\[\text{Power in hp} = \frac{490}{746} \approx 0.66 \, \text{hp}\]