A lift of mass 920 kg has a capacity of 10 persons. If average mass of person is 68 kg. Friction force between lift and lift shaft is 6000 N. The minimum power of motor required to move the lift upward with constant velocity 3 m/s is [`g = 10 m/s^2` ]

Options

A. 62360 W
B. 56300 W
C. 48000 W
D. 66000 W

Answer

To solve the problem, we need to find the minimum power required by the motor to move the lift upward with a constant velocity of 3 m/s. We will follow these steps:

### Step-by-Step Solution:

1. **Calculate the total weight of the lift and the passengers:**
- The mass of the lift (m_lift) = 920 kg
- The average mass of one person = 68 kg
- Maximum number of persons = 10
- Total mass of persons (m_persons) = 10 × 68 kg = 680 kg
- Total mass (m_total) = m_lift + m_persons = 920 kg + 680 kg = 1600 kg
- Weight (W) = m_total × g = 1600 kg × 10 m/s² = 16000 N

2. **Account for the friction force:**
- Given friction force (F_friction) = 6000 N

3. **Calculate the total force (tension) required to lift the system:**
- Since the lift is moving upward with constant velocity, the net force acting on the lift is zero.
- Therefore, the tension (T) in the cable must balance the weight of the lift and the friction force:
\[ T = W + F_{friction} = 16000 N + 6000 N = 22000 N \]

4. **Calculate the power required by the motor:**
- Power (P) is given by the formula:
\[ P = T \times v \]
- Where:
- T = tension = 22000 N
- v = velocity = 3 m/s
- Substituting the values:
\[ P = 22000 N \times 3 m/s = 66000 W \]

5. **Convert power to kilowatts (if needed):**
- Since 1 kW = 1000 W:
\[ P = \frac{66000 W}{1000} = 66 kW \]

### Final Answer:
The minimum power of the motor required to move the lift upward with a constant velocity of 3 m/s is **66000 W** or **66 kW**.