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 \( H \) using a rope of mass \( m \) and a bucket of mass \( M \), we can follow these steps:

### Step 1: Understand the System
We have a rope of mass \( m \) and a bucket of mass \( M \) at the bottom of a well of depth \( H \). When the bucket is pulled up, both the bucket and the rope will gain potential energy.

### Step 2: Calculate the Potential Energy Gained by the Bucket
The potential energy gained by the bucket when it is lifted to height \( H \) is given by the formula:
\[ PE_{\text{bucket}} = M \cdot g \cdot H \]
where \( g \) is the acceleration due to gravity.

### Step 3: Calculate the Potential Energy Gained by the Rope
The rope is uniform, and its center of mass is located at its midpoint. As the entire rope is lifted, the center of mass of the rope will rise by \( \frac{H}{2} \). Therefore, the potential energy gained by the rope is:
\[ PE_{\text{rope}} = m \cdot g \cdot \frac{H}{2} \]

### Step 4: Calculate the Total Work Done
The total work done by the person in pulling the bucket and the rope is the sum of the potential energies gained by both:
\[ W = PE_{\text{bucket}} + PE_{\text{rope}} \]
Substituting the expressions we found:
\[ W = M \cdot g \cdot H + m \cdot g \cdot \frac{H}{2} \]

### Step 5: Factor Out Common Terms
We can factor out \( gH \) from the total work done:
\[ W = gH \left( M + \frac{m}{2} \right) \]

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