A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a horizontal floor. The coefficient of friction between A and B is 0.2 while that between B and floor is 0.5. when a horizontal force of 25 N is applied on the block B. the force of friction between A and B is

Options

A. zero
B. 3.9 N
C. 5.0 N
D. 49 N

Explanation

To solve the problem, we need to find the force of friction between block A and block B when a horizontal force of 25 N is applied to block B.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of block A (m_A) = 2 kg
- Mass of block B (m_B) = 8 kg
- Coefficient of friction between A and B (μ_AB) = 0.2
- Coefficient of friction between B and the floor (μ_BF) = 0.5
- Applied force on block B (F) = 25 N

2. **Calculate the Normal Force (N) on Block A:**
The normal force acting on block A is equal to its weight since it is resting on block B.
\[ N = m_A \cdot g = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \]

3. **Calculate the Maximum Static Friction (f_max) between A and B:**
The maximum static friction force that can act between block A and block B is given by:
\[ f_{\text{max}} = \mu_{AB} \cdot N = 0.2 \cdot 19.6 \, \text{N} = 3.92 \, \text{N} \]

4. **Calculate the Maximum Static Friction (f_max) between B and the Floor:**
The normal force acting on block B is the sum of the weights of both blocks:
\[ N_B = (m_A + m_B) \cdot g = (2 \, \text{kg} + 8 \, \text{kg}) \cdot 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]
The maximum static friction force between block B and the floor is:
\[ f_{B} = \mu_{BF} \cdot N_B = 0.5 \cdot 98 \, \text{N} = 49 \, \text{N} \]

5. **Determine the Effect of the Applied Force:**
The applied force on block B is 25 N. Since the maximum static friction between B and the floor (49 N) is greater than the applied force (25 N), block B will not move.

6. **Determine the Friction Force between A and B:**
Since block B does not move, the friction force between A and B will be equal to the force required to prevent A from sliding. The applied force (25 N) is less than the maximum static friction (3.92 N). Therefore, the friction force between A and B will be equal to the applied force:
\[ f_{AB} = 25 \, \text{N} \]

### Conclusion:
The force of friction between block A and block B is **3.92 N**.