First, we determine if the block slips on the slab. If they move together, the system's acceleration would be $a_{sys} = \frac{F}{m_1 + m_2}$. The force required to accelerate the slab at this rate is provided by static friction, $f_{s,req} = m_1 a_{sys} = \frac{m_1 F}{m_1 + m_2}$.

The maximum available static friction force is $f_{s,max} = \mu_s N_2 = \mu_s m_2 g$. Slipping occurs if the required force exceeds the maximum available force, i.e., if $f_{s,req} > f_{s,max}$. This condition implies that the bodies move separately.

When the block slips, a kinetic friction force $f_k = \mu_k m_2 g$ acts. We apply Newton's second law to each body individually.

For the block ($m_2$), the net force is the applied force minus the opposing kinetic friction:

$$F - f_k = m_2 a_2 \implies a_2 = \frac{F - \mu_k m_2 g}{m_2}$$

For the slab ($m_1$), the only horizontal force is the kinetic friction exerted by the block, which acts in the direction of motion:

$$f_k = m_1 a_1 \implies a_1 = \frac{\mu_k m_2 g}{m_1}$$