An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.20 m/s. What would be the minimum horsepower of the motor to be used?

Options

A. `10.30 hp`
B. `5.15 hp`
C. 2.62 hp
D. 1.31 hp

Explanation

To solve the problem of determining the minimum horsepower required for a motor to lift an elevator weighing 500 kg at a constant velocity of 0.20 m/s, we can follow these steps:

### Step 1: Calculate the weight of the elevator
The weight (W) of the elevator can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 500 \, \text{kg} \) (mass of the elevator)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ W = 500 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 4900 \, \text{N} \]

### Step 2: Calculate the power required to lift the elevator
Power (P) can be calculated using the formula:
\[ P = F \cdot v \]
where:
- \( F \) is the force (which is equal to the weight in this case)
- \( v = 0.20 \, \text{m/s} \) (velocity)

Substituting the values:
\[ P = 4900 \, \text{N} \cdot 0.20 \, \text{m/s} = 980 \, \text{W} \]

### Step 3: Convert power from watts to horsepower
To convert power from watts to horsepower, we use the conversion factor:
\[ 1 \, \text{hp} = 746 \, \text{W} \]

Thus, the power in horsepower (hp) can be calculated as:
\[ P_{\text{hp}} = \frac{P}{746} \]
Substituting the power we calculated:
\[ P_{\text{hp}} = \frac{980 \, \text{W}}{746 \, \text{W/hp}} \approx 1.31 \, \text{hp} \]

### Final Answer
The minimum horsepower of the motor required is approximately:
\[ \text{Minimum horsepower} \approx 1.31 \, \text{hp} \]

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