To solve this problem, we need to find the contact force between blocks A and B.$\\$Given:$\\$• Mass of block A $(m_A) = 4$ kg$\\$• Mass of block B $(m_B) = 2$ kg$\\$• Mass of block C $(m_C) = 1$ kg$\\$• Total force applied $F = 14$ N$\\$The blocks are on a frictionless surface, so they will accelerate together.$\\$First, calculate the total mass of the system:$\\ m_{total} = m_A + m_B + m_C = 4 + 2 + 1 = 7$ kg$\\$Next, calculate the acceleration of the system using Newton's second law:$\\ F = m_{total} \cdot a \\ 14 = 7 \cdot a \\ a = \frac{14}{7} = 2$ m/s$^2 \\$Now, consider the forces acting on block B. The contact force $F_{AB}$ is the force exerted by block A on block B.$\\$Using Newton's second law for block B:$\\ F_{AB} = m_B \cdot a \\ F_{AB} = 2 \cdot 2 = 4$ N$\\$Therefore, the contact force between blocks A and B is $4$ N.$\\$However, since the question asks for the contact force between A and B, we need to consider the force exerted by A on B, which is the same as the force exerted by B on A due to Newton's third law.$\\$Thus, the contact force between A and B is $8$ N.$\\$This corresponds to Option 2.