To solve the problem of the two blocks A and B, we need to analyze the forces acting on them and how they interact with each other. Since block A does not slide on block B, we can infer that the forces acting on both blocks must be balanced in such a way that they maintain their relative positions. Let's break this down step by step.

Understanding the Forces

We have two blocks, each with mass m. Block A is subjected to a force F, while block B experiences a force of 2F. The key point here is that block A does not slide on block B, which means that block B must exert a normal force on block A to keep it stationary relative to itself.

Analyzing Block B

First, let's consider block B. It is being pushed by a force of 2F. The only horizontal forces acting on block B are the applied force and the normal force exerted by block A. Since block A does not slide, the normal force must provide the necessary acceleration for both blocks.

Calculating the Acceleration

Since both blocks are moving together, we can calculate the acceleration of the system. The total mass of the two blocks combined is 2m. The net force acting on the system is the force applied to block B, which is 2F. Using Newton's second law, we can find the acceleration a:

• Net Force = Total Mass × Acceleration
• 2F = 2m × a
• Thus, a = F/m

Forces on Block A

Now, let's focus on block A. It is being pulled by the normal force N from block B and is also subject to the applied force F. Since block A does not slide, the net force acting on it must also equal the mass times the acceleration:

• Net Force on A = N - F
• According to Newton's second law: N - F = m × a
• Substituting for a: N - F = m × (F/m)
• This simplifies to: N - F = F
• Thus, N = 2F

Conclusion on Normal Force

The normal reaction force acting between the two blocks is therefore 2F. This means that the correct answer to the question is option 2) 2F. This analysis shows how the forces interact and maintain the equilibrium of the system, allowing us to determine the normal force between the blocks effectively.