A bucket of mass `'m'` tied to a light rope is lowered at a constant acceleration of `g//4`. IF the bucket is lowered by a distance `'d'` , the work done by the rope will be (neglect the mass of the rope)

Options

A. `(1)/(4) mg d`
B. `(3)/(4) mg d`
C. `-(3)/(4) mg d`
D. `-(5)/(4)mg d`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the forces acting on the bucket
The forces acting on the bucket are:
1. The gravitational force (weight) acting downward, which is \( mg \).
2. The tension force \( T \) in the rope 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 use Newton's second law:
\[ F_{\text{net}} = ma \]
Here, the net force acting on the bucket is the difference between the gravitational force and the tension:
\[ mg - T = m \left( \frac{g}{4} \right) \]

### Step 3: Solve for the tension \( T \)
Rearranging the equation gives:
\[ T = mg - m \left( \frac{g}{4} \right) \]
\[ T = mg - \frac{mg}{4} \]
\[ T = \frac{4mg}{4} - \frac{mg}{4} = \frac{3mg}{4} \]

### Step 4: Calculate the work done by the rope
The work done by the tension in the rope when the bucket is lowered by a distance \( d \) can be calculated using the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
Where:
- \( F \) is the tension \( T = \frac{3mg}{4} \),
- \( d \) is the distance lowered,
- \( \theta \) is the angle between the force and the displacement. Since the tension acts upward and the displacement is downward, \( \theta = 180^\circ \).

Thus, \( \cos(180^\circ) = -1 \):
\[ W = \left(\frac{3mg}{4}\right) \cdot d \cdot (-1) \]
\[ W = -\frac{3mgd}{4} \]

### Final Answer
The work done by the rope when the bucket is lowered by a distance \( d \) is:
\[ W = -\frac{3mgd}{4} \]

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