To determine the accelerations of the blocks (and whether the top block slides), we'll need to analyze the forces acting on each block and apply Newton's second law of motion. Here's a step-by-step approach:

Step 1: Analyze forces on the top block ($m_2$)

• The applied force $\mathbf{P}$ has a horizontal component $P_x = P \cos(\theta) = 200 \cos(30^\circ)$ and a vertical component $P_y = P \sin(\theta) = 200 \sin(30^\circ)$.
• The gravitational force acting on the block is $F_{\text{gravity},2} = m_2 g = 20 \times 9.8 \, \text{N}$.
• The normal force $N_{2}$ exerted by the bottom block on the top block adjusts for $F_{\text{gravity},2}$ and the vertical component of $P$:
  $N_{2} = m_2 g - P_y = 20 \cdot 9.8 - 200 \cdot \sin(30^\circ)$

Step 2: Determine if the top block slides

• Calculate the maximum static friction force:
  $f_{\text{max}} = \mu_s N_{2}$

• Check if $P_x$ exceeds this maximum static friction force: $P_x = 200 \cos(30^\circ)$ If $P_x \le f_{\text{max}}$, the top block does not slide. Otherwise, it slides.

Step 3a: Top block does not slide

If the top block does not slide, both blocks move together with the same acceleration $a$:

• The friction force on $m_2$ acts to the left and there is no horizontal acceleration on $m_1$.

Using Newton's second law for both blocks together:

• Total mass $= m_1 + m_2 = 70 \, \text{kg}$

• Total force $= P_x$

  $a = \frac{P_x}{m_1 + m_2} = \frac{200 \cos(30^\circ)}{70}$

Step 3b: Top block slides

If the top block slides, it means $P_x > f_{\text{max}}$. In this case, the friction is kinetic friction.

The accelerations are separate:

For $m_2$:

• Kinetic friction force acting backward: $f_k = \mu_k N_2$

  $a_2 = \frac{P_x - f_k}{m_2}$

For $m_1$:

• Force due to friction acting forward: $f_k$

  $a_1 = \frac{f_k}{m_1}$

Conclusion

Calculations:
Using values, we find:

• $N_2 = 196 - 100 = 96 \, \text{N}$
• Maximum static friction $f_{\text{max}} = 0.6 \times 96 = 57.6 \, \text{N}$
• $P_x = 200 \cos(30^\circ) = 173.2 \, \text{N}$

Since $173.2 > 57.6$, the block slides.

Thus:

• Kinetic friction $f_k = 0.3 \times 96 = 28.8 \, \text{N}$
• $a_2 = \frac{173.2 - 28.8}{20} = 7.72 \, \text{m/s}^2$
• $a_1 = \frac{28.8}{50} = 0.576 \, \text{m/s}^2$