A man is drawing water from a well with a bucket which leaks uniformly. The bucket when full weights 20 kg and when it arrives the top only half the water remains. The depth of the water is 20 m What is the work done ?

Answer

To solve the problem of calculating the work done by a man drawing water from a well with a leaking bucket, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Initial Conditions**:
- The weight of the full bucket = 20 kg.
- The depth of the well = 20 m.
- When the bucket reaches the top, only half of the water remains, so the weight of the bucket at the top = 10 kg.

2. **Determine the Weight Loss**:
- Since the bucket loses half of its water, it loses 10 kg of weight over the distance of 20 m.
- The rate of weight loss per meter = Total weight loss / Depth of the well = 10 kg / 20 m = 0.5 kg/m.

3. **Define the Weight as a Function of Depth**:
- Let \( x \) be the depth from the top of the well (0 m at the top to 20 m at the bottom).
- The weight of the bucket at depth \( x \) can be expressed as:
\[ W(x) = 20 - 0.5x \quad \text{(in kg)} \]
- To convert this weight into force (in Newtons), we multiply by the acceleration due to gravity \( g \) (approximately \( 10 \, \text{m/s}^2 \)):
\[ F(x) = (20 - 0.5x) \cdot g = (20 - 0.5x) \cdot 10 = 200 - 5x \quad \text{(in N)} \]

4. **Set Up the Work Done Integral**:
- The work done \( W \) is given by the integral of the force over the distance from 0 to 20 m:
\[ W = \int_0^{20} F(x) \, dx = \int_0^{20} (200 - 5x) \, dx \]

5. **Calculate the Integral**:
- Now we can calculate the integral:
\[ W = \left[ 200x - \frac{5}{2}x^2 \right]_0^{20} \]
- Evaluating this from 0 to 20:
\[ W = \left[ 200(20) - \frac{5}{2}(20^2) \right] - \left[ 200(0) - \frac{5}{2}(0^2) \right] \]
\[ = 4000 - \frac{5}{2}(400) = 4000 - 1000 = 3000 \, \text{J} \]

6. **Final Result**:
- The total work done in drawing the water from the well is \( 3000 \, \text{J} \) or \( 3 \, \text{kJ} \).