To find the acceleration of the two blocks, we will follow these steps:
### Step 1: Identify the forces acting on the blocks
- We have two blocks: Block A (2 kg) and Block B (3 kg).
- A force of 5 N acts on Block A (to the right).
- A force of 10 N acts on Block B (to the right).
- The coefficient of friction between the blocks and the surface is given as 0.1.
### Step 2: Calculate the total external force
The total external force acting on the system of blocks is the sum of the forces acting on both blocks:
\[
F_{\text{external}} = F_A + F_B = 5 \, \text{N} + 10 \, \text{N} = 15 \, \text{N}
\]
### Step 3: Calculate the frictional force
The frictional force between the blocks and the surface must be calculated to determine the net force acting on the system. The normal force (N) acting on the blocks is the weight of both blocks combined:
\[
N = (m_A + m_B) \cdot g = (2 \, \text{kg} + 3 \, \text{kg}) \cdot 10 \, \text{m/s}^2 = 5 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 50 \, \text{N}
\]
The maximum static frictional force (F_friction) can be calculated as:
\[
F_{\text{friction}} = \mu \cdot N = 0.1 \cdot 50 \, \text{N} = 5 \, \text{N}
\]
### Step 4: Calculate the net force
The net force (F_net) acting on the system is given by the total external force minus the frictional force:
\[
F_{\text{net}} = F_{\text{external}} - F_{\text{friction}} = 15 \, \text{N} - 5 \, \text{N} = 10 \, \text{N}
\]
### Step 5: Calculate the total mass of the system
The total mass (m_total) of the system is the sum of the masses of both blocks:
\[
m_{\text{total}} = m_A + m_B = 2 \, \text{kg} + 3 \, \text{kg} = 5 \, \text{kg}
\]
### Step 6: Calculate the acceleration
Using Newton's second law, we can find the acceleration (a) of the system:
\[
F_{\text{net}} = m_{\text{total}} \cdot a \implies a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{10 \, \text{N}}{5 \, \text{kg}} = 2 \, \text{m/s}^2
\]
### Conclusion
The acceleration of the two blocks is:
\[
\boxed{2 \, \text{m/s}^2}
\]
