In the previous problem , if the applied force is made `20 N` , the acceleration of the upper and the lower blocks will be

Options

A. `4 m//s^(2) , 2 m//s^(2)`
B. `5 m//s^(2) , 2 m//s^(2)`
C. `5 m//s^(2) , 1 m//s^(2)`
D. `4 m//s^(2) , 1 m//s^(2)`

Explanation

To solve the problem, we need to determine the acceleration of the upper block (2 kg) and the lower block (5 kg) when an applied force of 20 N is exerted on the system.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Blocks:**
- The applied force \( F = 20 \, \text{N} \).
- The limiting friction force \( f_{\text{max}} = 10 \, \text{N} \) (as stated in the previous problem).
- The weight of the upper block (2 kg) is \( W_1 = m_1 \cdot g = 2 \cdot 10 = 20 \, \text{N} \).
- The weight of the lower block (5 kg) is \( W_2 = m_2 \cdot g = 5 \cdot 10 = 50 \, \text{N} \).

2. **Determine the Net Force on the System:**
- Since the applied force (20 N) is greater than the limiting friction (10 N), the blocks will slide together.
- The net force acting on the system is given by:
\[ F_{\text{net}} = F - f_{\text{max}} = 20 \, \text{N} - 10 \, \text{N} = 10 \, \text{N} \]

3. **Calculate the Total Mass of the System:**
- The total mass \( m_{\text{total}} = m_1 + m_2 = 2 \, \text{kg} + 5 \, \text{kg} = 7 \, \text{kg} \).

4. **Apply Newton's Second Law to Find the Acceleration:**
- According to Newton's second law, \( F = m \cdot a \).
- We can rearrange this to find acceleration \( a \):
\[ a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{10 \, \text{N}}{7 \, \text{kg}} \approx 1.43 \, \text{m/s}^2 \]

5. **Determine the Acceleration of Each Block:**
- Since both blocks are moving together, they will have the same acceleration.
- Therefore, the acceleration of both the upper block (2 kg) and the lower block (5 kg) is approximately \( 1.43 \, \text{m/s}^2 \).

### Final Answer:
- The acceleration of the upper block (2 kg) is \( 1.43 \, \text{m/s}^2 \).
- The acceleration of the lower block (5 kg) is also \( 1.43 \, \text{m/s}^2 \).