A block of mass 5 kg is placed on a plank of mass 10 kg which is placed on the smooth horizontal surface as shown in figure. If a 100 N force is applied on the plank then the acceleration of the plank wrt ground is (Coefficient of friction between block and plank `mu_s= mu_k=0.4` and `g = 10 m/s^2`)

Options

A. `8 = m/s ^2`
B. `2 = m/s ^2`
C. `6 = m/s ^2`
D. Zero

Answer

To solve the problem, we need to determine the acceleration of the plank with respect to the ground when a force of 100 N is applied to it. We will consider the forces acting on both the block and the plank, as well as the friction between them.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the block (m1) = 5 kg
- Mass of the plank (m2) = 10 kg
- Applied force (F) = 100 N
- Coefficient of friction (μ) = 0.4
- Acceleration due to gravity (g) = 10 m/s²

2. **Calculate the Maximum Static Friction (F_max):**
The maximum static friction that can act on the block is given by:
\[ F_{\text{max}} = \mu \cdot m_1 \cdot g \]
Substituting the values:
\[ F_{\text{max}} = 0.4 \cdot 5 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]

3. **Set Up the Equations of Motion:**
- For the block (5 kg):
The friction force (F_f) is what causes the block to accelerate. Thus, we can write:
\[ F_f = m_1 \cdot a \]
where \( a \) is the acceleration of the block.

- For the plank (10 kg):
The net force acting on the plank is the applied force minus the friction force:
\[ F - F_f = m_2 \cdot a \]
Substituting \( F_f \) from the block's equation:
\[ 100 - F_f = 10 \cdot a \]

4. **Combine the Equations:**
From the block's equation, we have:
\[ F_f = 5 \cdot a \]
Substituting this into the plank's equation:
\[ 100 - 5a = 10a \]
Rearranging gives:
\[ 100 = 15a \]
Therefore, solving for \( a \):
\[ a = \frac{100}{15} = \frac{20}{3} \approx 6.67 \, \text{m/s}^2 \]

5. **Check if the Block Slips:**
We need to check if the friction force required exceeds the maximum static friction:
\[ F_f = 5 \cdot a = 5 \cdot \frac{20}{3} = \frac{100}{3} \approx 33.33 \, \text{N} \]
Since \( 33.33 \, \text{N} > 20 \, \text{N} \), the block will slip.

6. **Calculate the Acceleration of the Plank:**
When slipping occurs, we use kinetic friction:
\[ F_{\text{friction}} = \mu_k \cdot m_1 \cdot g = 20 \, \text{N} \]
Now, the equation for the plank becomes:
\[ 100 - 20 = 10a \]
Simplifying gives:
\[ 80 = 10a \implies a = 8 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the plank with respect to the ground is **8 m/s²**.