A mass M is lowered with the help of a string by a distance h at a distance acceleration `g//2`. The work done by the string will be

Options

A. (a) `(Mgh)/(2)`
B. (b) `-(Mgh)/(2)`
C. (c) `(3Mgh)/(2)`
D. (d) `-(3Mgh)/(2)`

Explanation

To solve the problem, we need to find the work done by the string when a mass \( M \) is lowered by a distance \( h \) with an acceleration of \( \frac{g}{2} \).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Mass:**
- The weight of the mass \( M \) acting downwards: \( W = Mg \)
- The tension \( T \) in the string acting upwards.

2. **Set Up the Force Equation:**
- According to Newton's second law, the net force acting on the mass is equal to the mass times its acceleration.
- The acceleration of the mass is given as \( \frac{g}{2} \).
- Therefore, the equation can be written as:
\[ Mg - T = M \left(\frac{g}{2}\right) \]

3. **Solve for Tension \( T \):**
- Rearranging the equation gives:
\[ T = Mg - M\left(\frac{g}{2}\right) \]
- Simplifying this:
\[ T = Mg - \frac{Mg}{2} = \frac{Mg}{2} \]

4. **Calculate the Work Done by the String:**
- The work done by the tension in the string is given by the formula:
\[ W = T \cdot h \cdot \cos(\theta) \]
- Here, \( \theta \) is the angle between the tension force and the displacement. Since the tension acts upwards and the displacement is downwards, \( \theta = 180^\circ \).
- Therefore, \( \cos(180^\circ) = -1 \).
- Substituting \( T \) and \( \theta \) into the work formula:
\[ W = \left(\frac{Mg}{2}\right) \cdot h \cdot (-1) \]
- This simplifies to:
\[ W = -\frac{Mgh}{2} \]

5. **Final Answer:**
- The work done by the string is:
\[ W = -\frac{Mgh}{2} \]