A man slowly pulls a bucket of water from a well of depth `h=20m`. The mass of the uniform rope and bucket full of water are `m=200g` and M `19.9kg`, respectively. Find the work done `(in kJ)` by the mean.

Explanation

To find the work done by the man in pulling the bucket of water from the well, we can use the concept of gravitational potential energy. The work done is equal to the change in potential energy of both the bucket and the rope as they are lifted.

### Step-by-Step Solution:

1. **Identify the masses and height:**
- Mass of the bucket (full of water), \( M = 19.9 \, \text{kg} \)
- Mass of the rope, \( m = 200 \, \text{g} = 0.2 \, \text{kg} \) (convert grams to kilograms)
- Depth of the well, \( h = 20 \, \text{m} \)

2. **Calculate the change in potential energy for the bucket:**
- The potential energy change for the bucket when it is lifted from a depth of 20 m to the top is given by:
\[ \Delta PE_{\text{bucket}} = M \cdot g \cdot h \]
- Where \( g \approx 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).
- Substituting the values:
\[ \Delta PE_{\text{bucket}} = 19.9 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 20 \, \text{m} \]

3. **Calculate the change in potential energy for the rope:**
- The rope is uniform, and its center of mass is at half its length when fully extended. Therefore, the center of mass of the rope is at a height of \( \frac{20}{2} = 10 \, \text{m} \).
- The potential energy change for the rope is given by:
\[ \Delta PE_{\text{rope}} = m \cdot g \cdot h_{\text{cm}} \]
- Where \( h_{\text{cm}} = 10 \, \text{m} \) (height of the center of mass of the rope).
- Substituting the values:
\[ \Delta PE_{\text{rope}} = 0.2 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 10 \, \text{m} \]

4. **Calculate the total work done:**
- The total work done by the man is the sum of the work done to lift the bucket and the rope:
\[ W = \Delta PE_{\text{bucket}} + \Delta PE_{\text{rope}} \]

5. **Perform the calculations:**
- For the bucket:
\[ \Delta PE_{\text{bucket}} = 19.9 \cdot 9.81 \cdot 20 = 3900.78 \, \text{J} \]
- For the rope:
\[ \Delta PE_{\text{rope}} = 0.2 \cdot 9.81 \cdot 10 = 19.62 \, \text{J} \]
- Total work done:
\[ W = 3900.78 + 19.62 = 3920.4 \, \text{J} \]

6. **Convert the work done to kilojoules:**
\[ W = \frac{3920.4}{1000} = 3.9204 \, \text{kJ} \]

### Final Answer:
The work done by the man is approximately \( 3.92 \, \text{kJ} \).