Consider the situation as shown in the diagram.
The coefficient of friction between the block is `0.5`. The acceleration of the lower block is

Options

A. `1 m//s^(2)`
B. `1.5 m//s^(2)`
C. `2 m//s^(2)`
D. `2.5 m//s^(2)`

Explanation

To solve the problem step by step, we will analyze the forces acting on the blocks and calculate the acceleration of the lower block (5 kg) based on the given conditions.

### Step 1: Identify the Forces
We have two blocks:
- Block 1 (2 kg) is being pulled by a force of 7 N.
- Block 2 (5 kg) is on the ground and experiences friction from Block 1.

The coefficient of friction (μ) between the blocks is 0.5.

### Step 2: Calculate the Normal Force
The normal force (N) acting on Block 1 (2 kg) due to gravity is:
\[ N = m_1 \cdot g = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]

### Step 3: Calculate the Limiting Friction
The limiting friction (f) between the two blocks can be calculated using the formula:
\[ f_{\text{max}} = \mu \cdot N \]
Substituting the values:
\[ f_{\text{max}} = 0.5 \cdot 20 \, \text{N} = 10 \, \text{N} \]

### Step 4: Compare the Applied Force with Limiting Friction
The applied force on Block 1 is 7 N, which is less than the maximum limiting friction of 10 N. Therefore, the frictional force will not be overcome, and both blocks will move together without slipping.

### Step 5: Calculate the Acceleration of the System
Since there is no relative motion between the blocks, we can treat them as a single system. The total mass of the system is:
\[ m_{\text{total}} = m_1 + m_2 = 2 \, \text{kg} + 5 \, \text{kg} = 7 \, \text{kg} \]

Using Newton's second law, the acceleration (a) of the system can be calculated as:
\[ F = m_{\text{total}} \cdot a \]
Rearranging gives:
\[ a = \frac{F}{m_{\text{total}}} = \frac{7 \, \text{N}}{7 \, \text{kg}} = 1 \, \text{m/s}^2 \]

### Step 6: Conclusion
The acceleration of the lower block (5 kg) is:
\[ a = 1 \, \text{m/s}^2 \]