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. 2m/s
B. 2.9 m/s
C. 1.7 m/s
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. **Calculate the Weight of the Elevator:**
The weight \( W \) of the elevator can be calculated using the formula:
\[ W = m \cdot g \]
where \( m = 3000 \, \text{kg} \) (mass of the elevator) and \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
\[ W = 3000 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 30000 \, \text{N} \]

2. **Calculate the Total Force Required to Lift the Elevator:**
The total force \( F \) required to lift the elevator must overcome both the weight of the elevator and the frictional force \( F_f \):
\[ F = W + F_f \]
where \( F_f = 6000 \, \text{N} \) (frictional force).
\[ F = 30000 \, \text{N} + 6000 \, \text{N} = 36000 \, \text{N} \]

3. **Calculate the Power of the Motor:**
The power \( P \) of the motor is given as 120 HP. To convert this to watts, we use the conversion factor \( 1 \, \text{HP} = 746 \, \text{W} \):
\[ P = 120 \, \text{HP} \cdot 746 \, \text{W/HP} = 89520 \, \text{W} \]

4. **Relate Power to Force and Velocity:**
The power can also be expressed in terms of force and velocity as:
\[ P = F \cdot v \]
Rearranging this gives us the velocity \( v \):
\[ v = \frac{P}{F} \]

5. **Substitute the Values:**
Now we can substitute the values we calculated:
\[ v = \frac{89520 \, \text{W}}{36000 \, \text{N}} \approx 2.49 \, \text{m/s} \]

6. **Final Result:**
The speed of the elevator at full load is approximately \( 2.5 \, \text{m/s} \).