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 `l` then find the work done by the string on the bucket.

Options

A. `-3/4mgl`
B. `3/4mgl`
C. `4/3mgl`
D. `-4/3mgl`

Explanation

To find the work done by the string on the bucket, we can follow these steps:

### Step 1: Identify the forces acting on the bucket
The forces acting on the bucket are:
- The weight of the bucket acting downwards, which is \( mg \).
- The tension in the string acting upwards, which we will denote as \( T \).

### Step 2: Write the equation of motion
Since the bucket is being lowered with a constant acceleration of \( \frac{g}{4} \), we can apply Newton's second law:
\[ mg - T = ma \]
Here, \( a = \frac{g}{4} \). Substituting this into the equation gives:
\[ mg - T = m \left(\frac{g}{4}\right) \]

### Step 3: Solve for the tension \( T \)
Rearranging the equation to isolate \( T \):
\[ T = mg - m \left(\frac{g}{4}\right) \]
Factoring out \( mg \):
\[ T = mg \left(1 - \frac{1}{4}\right) = mg \left(\frac{3}{4}\right) \]
Thus, the tension \( T \) is:
\[ T = \frac{3}{4} mg \]

### Step 4: Calculate the work done by the string
The work done by the string on the bucket is given by:
\[ W = T \cdot s \cdot \cos(\theta) \]
Where:
- \( T \) is the tension we found,
- \( s \) is the distance the bucket is lowered, which is \( l \),
- \( \theta \) is the angle between the tension and the displacement. Since tension acts upwards and the displacement is downwards, \( \theta = 180^\circ \) (or \( \pi \) radians), and \( \cos(180^\circ) = -1 \).

Substituting the values:
\[ W = \left(\frac{3}{4} mg\right) \cdot l \cdot (-1) \]
This simplifies to:
\[ W = -\frac{3}{4} mgl \]

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

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