A book of weight `20N` is pressed between two hands and each hand exerts a force of `40N`. If the block just starts to slide down Coefficient of friction is .

Options

A. `0.25`
B. `0.2`
C. `0.5`
D. `0.1`

Explanation

To solve the problem, we need to determine the coefficient of friction when a book weighing 20 N is pressed between two hands, each exerting a force of 40 N, and the book just starts to slide down.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Book:**
- The weight of the book (W) is given as 20 N.
- Each hand exerts a force (F) of 40 N, resulting in a total normal force (N) acting on the book from both hands.

2. **Calculate the Total Normal Force:**
- Since there are two hands, the total normal force acting on the book is:
\[ N = F_{1} + F_{2} = 40 N + 40 N = 80 N \]

3. **Understand the Frictional Forces:**
- The frictional force (f) that opposes the motion of the book when it starts sliding is given by:
\[ f = \mu \cdot N \]
- Here, \(\mu\) is the coefficient of friction.

4. **Set Up the Equation for Sliding:**
- When the book just starts to slide down, the total frictional force must equal the weight of the book:
\[ f = W \]
- Thus, we have:
\[ \mu \cdot N = W \]

5. **Substituting Known Values:**
- Substitute \(N = 80 N\) and \(W = 20 N\) into the equation:
\[ \mu \cdot 80 N = 20 N \]

6. **Solve for the Coefficient of Friction:**
- Rearranging the equation gives:
\[ \mu = \frac{20 N}{80 N} = \frac{1}{4} = 0.25 \]

### Conclusion:
The coefficient of friction (\(\mu\)) is 0.25.