An elevator can carry a maximum load of `1800 kg` (elevator + passengers) is moving up with a constant speed of `2 ms^(-1)`. The frictional force opposing the motion is `4000 N`. What is minimum power delivered by the motor to the elevator?

Options

A. `22 kW`
B. `44 kW`
C. `66 kW`
D. `88 kW`

Explanation

To find the minimum power delivered by the motor to the elevator, we can follow these steps:

### Step 1: Identify the forces acting on the elevator
The elevator is moving upwards with a constant speed, which means the net force acting on it is zero. The forces acting on the elevator include:
- The gravitational force (weight) acting downwards, which is given by \( Mg \).
- The frictional force acting downwards, which is given as \( 4000 \, \text{N} \).

### Step 2: Calculate the weight of the elevator
The weight of the elevator can be calculated using the formula:
\[ Mg = m \cdot g \]
where:
- \( m = 1800 \, \text{kg} \) (mass of the elevator and passengers)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ Mg = 1800 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 17658 \, \text{N} \]

### Step 3: Calculate the total force opposing the motion
The total force opposing the upward motion of the elevator is the sum of the weight and the frictional force:
\[ F_{\text{total}} = Mg + F_{\text{friction}} = 17658 \, \text{N} + 4000 \, \text{N} = 21658 \, \text{N} \]

### Step 4: Calculate the power delivered by the motor
Power can be calculated using the formula:
\[ P = F \cdot v \]
where:
- \( F \) is the total force opposing the motion (calculated in Step 3)
- \( v = 2 \, \text{m/s} \) (constant speed of the elevator)

Substituting the values:
\[ P = 21658 \, \text{N} \cdot 2 \, \text{m/s} = 43316 \, \text{W} \]

### Step 5: Convert power to kilowatts
To convert watts to kilowatts, divide by 1000:
\[ P = \frac{43316 \, \text{W}}{1000} = 43.316 \, \text{kW} \]

### Final Answer
The minimum power delivered by the motor to the elevator is approximately \( 43.32 \, \text{kW} \).

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