A 120 HP electric motor lifts an elevator having a maximum total load capacity of 3000 kg. If the frictional force on the elevator is 6000N, the speed of the elevator at full load is close to : (1 HP = 746 `W, g = 10 ms^(-2)`).

Options

A. (a)2m/s
B. (b)2.9m/s
C. (c)1.8 m/s
D. (d)2.5 m/s

Explanation

To solve the problem, we need to determine the speed of the elevator when it is fully loaded, taking into account the power of the motor, the weight of the elevator, and the frictional force acting against it.

### Step-by-Step Solution:

1. **Convert Horsepower to Watts:**
Given that 1 HP = 746 W, we can convert the motor's power from horsepower to watts.
\[ \text{Power of motor} = 120 \text{ HP} \times 746 \text{ W/HP} = 89520 \text{ W} \]

2. **Calculate the Weight of the Elevator:**
The weight (force due to gravity) of the elevator can be calculated using the formula:
\[ \text{Weight} (W) = m \cdot g \]
where \( m = 3000 \text{ kg} \) and \( g = 10 \text{ m/s}^2 \).
\[ W = 3000 \text{ kg} \times 10 \text{ m/s}^2 = 30000 \text{ N} \]

3. **Determine the Total Force Acting on the Elevator:**
The total force that the motor needs to overcome includes both the weight of the elevator and the frictional force.
\[ \text{Total Force} = W + \text{Friction} = 30000 \text{ N} + 6000 \text{ N} = 36000 \text{ N} \]

4. **Relate Power, Force, and Velocity:**
The power provided by the motor is used to lift the elevator against the total force. The relationship between power, force, and velocity is given by:
\[ P = F \cdot v \]
Rearranging this gives us:
\[ v = \frac{P}{F} \]

5. **Substituting the Values:**
Now we can substitute the values of power and total force into the equation to find the velocity:
\[ v = \frac{89520 \text{ W}}{36000 \text{ N}} \approx 2.49 \text{ m/s} \]

6. **Final Result:**
Rounding to one decimal place, the speed of the elevator at full load is approximately:
\[ v \approx 2.5 \text{ m/s} \]

### Conclusion:
The speed of the elevator at full load is close to **2.5 m/s**.

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