Two bodies $A$ and $B$ of masses $5 \, kg$ and $10 \, kg$ in contact with each other rest on a table against a rigid wall. The coefficient of friction between the bodies and the table is $0.15$. $A$ force of $200 \, N$ is applied horizontally to $A$. What are $(a)$ the reaction of the partition $(b)$ the action-reaction forces between $A$ and $B$? What happens when the wall is removed? Does the answer to $(b)$ change,when the bodies are in motion? Ignore the difference between $\mu_{s}$ and $\mu_{k}$.

(N/A) Mass of body $A$,$m_{A} = 5 \, kg$.
Mass of body $B$,$m_{B} = 10 \, kg$.
Applied force,$F = 200 \, N$.
Coefficient of friction,$\mu = 0.15$.
The total force of friction is given by $f_{s} = \mu(m_{A} + m_{B})g = 0.15(5 + 10) \times 10 = 22.5 \, N$ (leftward).
$(a)$ The net force acting on the partition is $F_{net} = F - f_{s} = 200 - 22.5 = 177.5 \, N$ (rightward). By Newton's third law,the reaction of the wall is $177.5 \, N$ (leftward).
$(b)$ The force of friction on mass $A$ is $f_{A} = \mu m_{A} g = 0.15 \times 5 \times 10 = 7.5 \, N$ (leftward). The net force exerted by $A$ on $B$ is $F_{AB} = F - f_{A} = 200 - 7.5 = 192.5 \, N$ (rightward). By Newton's third law,$B$ exerts $192.5 \, N$ on $A$ (leftward).
When the wall is removed,the system moves with acceleration $a = \frac{F - f_{s}}{m_{A} + m_{B}} = \frac{177.5}{15} \approx 11.83 \, m/s^{2}$.
The force exerted by $A$ on $B$ becomes $F'_{AB} = m_{B} a + f_{B} = 10 \times 11.83 + (0.15 \times 10 \times 10) = 118.3 + 15 = 133.3 \, N$. Yes,the answer to $(b)$ changes when the bodies are in motion.