A bucket tied to a string is lowered at a constant acceleration of `g//4`. If mass of the bucket is m and it is lowered by a distance /then find then work done by the string on the bucket.

Explanation

To solve the problem of finding the work done by the string on the bucket, we can follow these steps:

### Step 1: Understand the forces acting on the bucket
The forces acting on the bucket are:
- The weight of the bucket, which is \( Mg \) (acting downward).
- The tension in the string, which we will denote as \( T \) (acting upward).

### Step 2: Apply Newton's second law
Since the bucket is being lowered with a constant acceleration of \( \frac{g}{4} \), we can apply Newton's second law:
\[ \text{Net force} = \text{mass} \times \text{acceleration} \]
This gives us:
\[ Mg - T = Ma \]
Where \( a = \frac{g}{4} \).

### Step 3: Substitute the value of acceleration
Substituting \( a \) into the equation:
\[ Mg - T = M \left( \frac{g}{4} \right) \]

### Step 4: Rearranging the equation to find tension
Rearranging the equation to solve for \( T \):
\[ T = Mg - M \left( \frac{g}{4} \right) \]
Factoring out \( M \) gives:
\[ T = M \left( g - \frac{g}{4} \right) = M \left( \frac{4g}{4} - \frac{g}{4} \right) = M \left( \frac{3g}{4} \right) \]
Thus, the tension \( T \) is:
\[ T = \frac{3Mg}{4} \]

### Step 5: Calculate the work done by the tension
The work done by the tension on the bucket can be calculated using the formula:
\[ \text{Work} = \text{Force} \times \text{Displacement} \times \cos(\theta) \]
Where:
- Force is the tension \( T \).
- Displacement is the distance \( L \) (the distance the bucket is lowered).
- \( \theta \) is the angle between the force and the displacement. Since the tension acts upward and the displacement is downward, \( \theta = 180^\circ \) and \( \cos(180^\circ) = -1 \).

Substituting these values gives:
\[ \text{Work} = T \times L \times \cos(180^\circ) = \left( \frac{3Mg}{4} \right) \times L \times (-1) \]
Thus:
\[ \text{Work} = -\frac{3MgL}{4} \]

### Final Answer
The work done by the string on the bucket is:
\[ \text{Work} = -\frac{3MgL}{4} \]

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