A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity.The frictional force between man and pole is equal to (in terms of man's weight `W`)

Options

A. `( w)/( 4)`
B. `( w)/( 2)`
C. `( 3w)/( 4)`
D. w

Answer

To solve the problem, we need to analyze the forces acting on the man as he slides down the telegraphic pole.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man:**
- The weight of the man (W) acts downward, which can be expressed as \( W = mg \) where \( m \) is the mass of the man and \( g \) is the acceleration due to gravity.
- There is a frictional force (F_f) acting upward, opposing the motion of the man.
- The man is accelerating downwards with an acceleration \( a = \frac{g}{4} \).

2. **Apply Newton's Second Law:**
- According to Newton's second law, the net force acting on the man is equal to the mass of the man multiplied by his acceleration:
\[ F_{\text{net}} = ma \]

3. **Set Up the Equation:**
- The net force acting on the man can be expressed as:
\[ F_{\text{net}} = W - F_f \]
- Substituting for \( F_{\text{net}} \):
\[ W - F_f = ma \]
- Since \( a = \frac{g}{4} \), we can express \( ma \) as:
\[ ma = m \cdot \frac{g}{4} \]

4. **Substituting Weight:**
- We know \( W = mg \), so we can rewrite the equation:
\[ W - F_f = \frac{W}{4} \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ F_f = W - \frac{W}{4} \]
- This simplifies to:
\[ F_f = \frac{4W}{4} - \frac{W}{4} = \frac{3W}{4} \]

6. **Conclusion:**
- Therefore, the frictional force \( F_f \) between the man and the pole is:
\[ F_f = \frac{3W}{4} \]

### Final Answer:
The frictional force between the man and the pole is \( \frac{3W}{4} \).