To solve this, we use the condition for no slipping between the two blocks.

1. Interpreting the figure/data

The standard configuration for this question is:

• a small block of mass $m_1 = 1\,\text{kg}$ placed on a larger block of mass $m_2 = 2\,\text{kg}$,
• coefficient of static friction between the blocks: $\mu_s = 0.5$,
• table is smooth, so no friction with ground,
• horizontal force $F$ is applied on the lower block,
• both blocks should move together.

This is consistent with the stored answer.

2. Maximum acceleration for moving together

For the upper block to move together with the lower block, friction must provide its acceleration.

Maximum static friction is $f_{\max} = \mu_s N = \mu_s m_1 g$

Since the only horizontal force on the top block is friction, $m_1 a \le f_{\max}$

So, $m_1 a \le \mu_s m_1 g$ $a \le \mu_s g$

Thus the maximum allowed acceleration is $a_{\max} = 0.5 \times 10 = 5\,\text{m/s}^2$

3. Relating force and acceleration of the whole system

If both blocks move together, total mass is $m_1 + m_2 = 1+2=3\,\text{kg}$

Hence, $F = (m_1+m_2)a$

For maximum force without slipping, $F_{\max} = (m_1+m_2)a_{\max}$ $F_{\max} = 3 \times 5 = 15\,\text{N}$

4. Final answer

$\boxed{15}$