A bucket fillled with water weighing 20 kg is raised from a well of depth 20m. If the linear density of the rope is 0.2 kg per meter, the amount of the work done is `(g=10ms^(-2))`

Options

A. 4000J
B. 4040J
C. 4400J
D. 4200J

Answer

To solve the problem of the work done in raising a bucket filled with water from a well, we need to consider both the weight of the bucket and the weight of the rope being lifted. Here is a step-by-step solution:

### Step 1: Understand the Problem
We have a bucket weighing 20 kg filled with water, and it is raised from a depth of 20 m. The rope has a linear density of 0.2 kg/m. We need to calculate the total work done in lifting the bucket and the rope.

### Step 2: Calculate the Force
The force required to lift the bucket and the rope can be expressed as:
\[ F = (m_{\text{bucket}} + m_{\text{rope}}) \cdot g \]
Where:
- \( m_{\text{bucket}} = 20 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( m_{\text{rope}} \) varies as we lift the rope.

### Step 3: Determine the Mass of the Rope
The mass of the rope lifted when the bucket is raised by a distance \( x \) is:
\[ m_{\text{rope}} = \text{linear density} \times \text{length of rope lifted} = 0.2 \, \text{kg/m} \times x \]

### Step 4: Write the Total Mass as a Function of x
Thus, the total mass being lifted when the bucket is at height \( x \) is:
\[ m_{\text{total}} = 20 + 0.2x \]

### Step 5: Calculate the Force as a Function of x
The force can thus be expressed as:
\[ F(x) = (20 + 0.2x) \cdot 10 = 200 + 2x \]

### Step 6: Set Up the Work Integral
The work done \( W \) in lifting the bucket from the bottom of the well (0 m) to the top (20 m) is given by the integral of the force over the distance:
\[ W = \int_{0}^{20} F(x) \, dx = \int_{0}^{20} (200 + 2x) \, dx \]

### Step 7: Evaluate the Integral
Now, we can compute the integral:
\[ W = \int_{0}^{20} (200 + 2x) \, dx = \int_{0}^{20} 200 \, dx + \int_{0}^{20} 2x \, dx \]
Calculating each part:
1. \( \int_{0}^{20} 200 \, dx = 200x \Big|_{0}^{20} = 200 \times 20 - 200 \times 0 = 4000 \)
2. \( \int_{0}^{20} 2x \, dx = 2 \cdot \frac{x^2}{2} \Big|_{0}^{20} = x^2 \Big|_{0}^{20} = 20^2 - 0^2 = 400 \)

### Step 8: Combine the Results
Now, add the results from both integrals:
\[ W = 4000 + 400 = 4400 \, \text{Joules} \]

### Final Answer
The total work done in raising the bucket filled with water from the well is:
\[ \boxed{4400 \, \text{Joules}} \]

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