A body of mass m is allowed to fall with the help of string with downward acceleration `(g)/(6)` to a distance x. The work done by the string is

Options

A. `(mgx)/(6)`
B. `-(mgx)/(6)`
C. `(5mgx)/(6)`
D. `-(5mgx)/(6)`

Answer

To solve the problem of finding the work done by the string when a body of mass \( m \) is allowed to fall with a downward acceleration of \( \frac{g}{6} \) over a distance \( x \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body**:
- The weight of the body acting downwards: \( W = mg \)
- The tension in the string acting upwards: \( T \)

2. **Apply Newton's Second Law**:
- According to Newton's second law, the net force acting on the body is equal to the mass of the body multiplied by its acceleration:
\[ F_{\text{net}} = m \cdot a \]
- Here, the acceleration \( a \) is given as \( \frac{g}{6} \). The net force can be expressed as:
\[ F_{\text{net}} = mg - T \]
- Setting these equal gives:
\[ mg - T = m \cdot \frac{g}{6} \]

3. **Solve for the Tension \( T \)**:
- Rearranging the equation:
\[ T = mg - m \cdot \frac{g}{6} \]
- Factor out \( mg \):
\[ T = mg \left(1 - \frac{1}{6}\right) = mg \cdot \frac{5}{6} \]

4. **Calculate the Work Done by the String**:
- The work done by the string is given by the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
- Here, \( F \) is the tension \( T \), \( d \) is the distance \( x \), and \( \theta \) is the angle between the force and the displacement. Since the tension acts upwards and the displacement is downwards, \( \theta = 180^\circ \) and \( \cos(180^\circ) = -1 \):
\[ W = T \cdot x \cdot \cos(180^\circ) = T \cdot x \cdot (-1) \]
- Substituting the value of \( T \):
\[ W = -\left(\frac{5mg}{6}\right) \cdot x \]

5. **Final Expression for Work Done**:
- Thus, the work done by the string is:
\[ W = -\frac{5mgx}{6} \]

### Final Answer:
The work done by the string is \( W = -\frac{5mgx}{6} \).