A lift weighing 250 kg is to be lifted up at a constant velocity of 0.20 m. What would be the minimum horse power of the motor to be used ?

Options

A. 1.3 hp
B. 0.65 hp
C. 1.5 hp
D. 0.75 hp

Answer

To solve the problem of determining the minimum horsepower of the motor required to lift a lift weighing 250 kg at a constant velocity of 0.20 m/s, we can follow these steps:

### Step 1: Calculate the Force Required to Lift the Lift
The force required to lift the lift can be calculated using the formula:
\[ F = m \cdot g \]
where:
- \( m \) is the mass of the lift (250 kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

Substituting the values:
\[ F = 250 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 2450 \, \text{N} \]

### Step 2: Calculate the Power Required
Power can be calculated using the formula:
\[ P = F \cdot v \]
where:
- \( P \) is the power,
- \( F \) is the force calculated in Step 1 (2450 N),
- \( v \) is the velocity (0.20 m/s).

Substituting the values:
\[ P = 2450 \, \text{N} \cdot 0.20 \, \text{m/s} = 490 \, \text{W} \]

### Step 3: Convert Power from Watts to Horsepower
To convert power from watts to horsepower, we use the conversion factor:
\[ 1 \, \text{horsepower} = 745.7 \, \text{W} \]

Thus, to convert 490 W to horsepower:
\[ \text{Horsepower} = \frac{490 \, \text{W}}{745.7 \, \text{W/hp}} \approx 0.656 \, \text{hp} \]

### Step 4: Round the Result
Rounding the result to two decimal places, we find:
\[ \text{Horsepower} \approx 0.66 \, \text{hp} \]

### Final Answer
The minimum horsepower of the motor required is approximately **0.66 hp**.

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