A cord is used to lower vertically a block of mass `M`, a distance `d` at a constant downward acceleration of `(g)/(4)`, then the work done by the cord on the block is

Options

A. `Mg(d)/(4)`
B. `3Mg(d)/(4)`
C. `-3Mg(d)/(4)`
D. `Mgd`

Explanation

To solve the problem, we need to determine the work done by the cord on the block as it is lowered at a constant downward acceleration.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The weight of the block (downward force) is given by \( W = mg \), where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity.
- The tension \( T \) in the cord acts upward.

2. **Apply Newton's Second Law:**
- Since the block is accelerating downward with an acceleration of \( \frac{g}{4} \), we can write the equation of motion as:
\[ mg - T = ma \]
- Here, \( a = \frac{g}{4} \). Substituting this into the equation gives:
\[ mg - T = m \left(\frac{g}{4}\right) \]

3. **Rearranging to Find Tension:**
- Rearranging the equation:
\[ T = mg - m\left(\frac{g}{4}\right) \]
- Simplifying this:
\[ T = mg - \frac{mg}{4} = mg\left(1 - \frac{1}{4}\right) = mg\left(\frac{3}{4}\right) \]
- Thus, the tension \( T \) is:
\[ T = \frac{3mg}{4} \]

4. **Calculate the Work Done by the Tension:**
- The work done \( W \) by the tension when the block is lowered a distance \( d \) is given by:
\[ W = T \cdot d \cdot \cos(\theta) \]
- Here, \( \theta = 180^\circ \) because the tension acts upward while the displacement is downward. Thus, \( \cos(180^\circ) = -1 \).
- Substituting the values:
\[ W = \left(\frac{3mg}{4}\right) \cdot d \cdot (-1) = -\frac{3mgd}{4} \]

5. **Final Result:**
- The work done by the cord on the block is:
\[ W = -\frac{3mgd}{4} \]

### Final Answer:
The work done by the cord on the block is \( -\frac{3mgd}{4} \).

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