A bucket full of water weighs 5 kg, it is pulled from a well 20 m deep. There is a small hole in the bucket through which water leaks at a constant rate. If it is observed that for every meter the bucket loses 0.2 kg mass of water, then the total work done in pulling the bucket up from the well is `[g = 10ms^(-2)`]

Options

A. 600 J
B. 400 J
C. 100 J
D. 500 J

Answer

To solve the problem of calculating the total work done in pulling the bucket up from the well, we can follow these steps:

### Step 1: Understand the Problem
The bucket weighs 5 kg and is being pulled up from a depth of 20 m. As it is pulled up, it loses water at a rate of 0.2 kg for every meter.

### Step 2: Determine the Mass of the Bucket as a Function of Height
Let \( x \) be the height in meters that the bucket has been pulled up. The mass of the water lost after pulling the bucket up \( x \) meters is given by:
\[ \text{Mass lost} = 0.2 \, \text{kg/m} \times x \, \text{m} = 0.2x \, \text{kg} \]
Thus, the mass of the bucket at height \( x \) is:
\[ m(x) = 5 \, \text{kg} - 0.2x \, \text{kg} \]

### Step 3: Calculate the Force Acting on the Bucket
The force due to gravity acting on the bucket as it is pulled up is:
\[ F(x) = m(x) \cdot g = (5 - 0.2x) \cdot 10 \]
where \( g = 10 \, \text{m/s}^2 \).

### Step 4: Set Up the Work Integral
The work done \( W \) in pulling the bucket from the bottom of the well (0 m) to the top (20 m) can be calculated using the integral of the force over the distance:
\[ W = \int_0^{20} F(x) \, dx = \int_0^{20} (5 - 0.2x) \cdot 10 \, dx \]
This simplifies to:
\[ W = 10 \int_0^{20} (5 - 0.2x) \, dx \]

### Step 5: Evaluate the Integral
Now we evaluate the integral:
\[ W = 10 \left[ 5x - 0.1x^2 \right]_0^{20} \]
Calculating the bounds:
\[ = 10 \left[ (5 \cdot 20) - 0.1 \cdot (20^2) \right] \]
\[ = 10 \left[ 100 - 0.1 \cdot 400 \right] \]
\[ = 10 \left[ 100 - 40 \right] = 10 \times 60 = 600 \, \text{J} \]

### Final Answer
The total work done in pulling the bucket up from the well is:
\[ \boxed{600 \, \text{J}} \]

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