A person pulls a bucket of water from a well of depth h. if the mass of uniform rope is and that of the bucket full 0f water is M, then work done by the person is.

Options

A. `(M + (m)/(2))gh`
B. 1/2 (M + m)gh`
C. `(M + m)gh`
D. `(M/2 + m)gh`

Explanation

To solve the problem of calculating the work done by a person pulling a bucket of water from a well of depth \( h \) with a uniform rope, we can break it down into the following steps:

### Step-by-Step Solution:

1. **Identify the Components**:
- Let the mass of the bucket full of water be \( M \).
- Let the mass of the rope be \( m \).
- The depth of the well is \( h \).

2. **Determine the Work Done on the Bucket**:
- The bucket is pulled from a depth \( h \) to the surface. The work done against gravity on the bucket is given by:
\[ W_{\text{bucket}} = M \cdot g \cdot h \]
where \( g \) is the acceleration due to gravity.

3. **Determine the Work Done on the Rope**:
- The rope has a uniform mass distribution. The center of mass of the rope will rise by \( \frac{h}{2} \) as the entire rope is pulled up.
- The work done against gravity on the rope is given by:
\[ W_{\text{rope}} = m \cdot g \cdot \frac{h}{2} \]

4. **Calculate the Total Work Done**:
- The total work done by the person is the sum of the work done on the bucket and the work done on the rope:
\[ W_{\text{total}} = W_{\text{bucket}} + W_{\text{rope}} \]
- Substituting the expressions we found:
\[ W_{\text{total}} = M \cdot g \cdot h + m \cdot g \cdot \frac{h}{2} \]

5. **Factor Out Common Terms**:
- We can factor out \( g \) and \( h \) from the total work:
\[ W_{\text{total}} = g \cdot h \left( M + \frac{m}{2} \right) \]

### Final Answer:
The work done by the person in pulling the bucket of water from the well is:
\[ W = g \cdot h \left( M + \frac{m}{2} \right) \]