An elevator can carry a maximum load of `1800 kg` (elevator + passengers) is moving up with a constant speed of `2 ms^(-1)`. The friction force opposite 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 solve the problem, we need to determine the minimum power delivered by the motor to the elevator. Here are the steps to find the solution:

### Step 1: Identify the given values
- Maximum load (mass of the elevator + passengers), \( m = 1800 \, \text{kg} \)
- Constant speed of the elevator, \( v = 2 \, \text{m/s} \)
- Frictional force opposing the motion, \( f = 4000 \, \text{N} \)

### Step 2: Calculate the gravitational force acting on the elevator
The gravitational force (\( F_g \)) can be calculated using the formula:
\[ F_g = m \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).

Substituting the values:
\[ F_g = 1800 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 18000 \, \text{N} \]

### Step 3: Calculate the total downward force on the elevator
The total downward force (\( F \)) acting on the elevator is the sum of the gravitational force and the frictional force:
\[ F = F_g + f \]
Substituting the values:
\[ F = 18000 \, \text{N} + 4000 \, \text{N} = 22000 \, \text{N} \]

### Step 4: Calculate the power delivered by the motor
The power (\( P \)) delivered by the motor can be calculated using the formula:
\[ P = F \cdot v \]
Substituting the values:
\[ P = 22000 \, \text{N} \cdot 2 \, \text{m/s} = 44000 \, \text{W} \]

### Step 5: Convert power to kilowatts
Since power is often expressed in kilowatts (kW), we convert it:
\[ P = \frac{44000 \, \text{W}}{1000} = 44 \, \text{kW} \]

### Final Answer
The minimum power delivered by the motor to the elevator is **44 kW**.

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