Two blocks A and B of mass mA and mB respectively are kept in contact on a frictionless table .yhe experimenter pushes the block A from behind so that the block accelerate .if the block A exerys a force F on the block B what is the force exerted by the experimenter on A

Question

Answer

Force exerted by the experimenter on Block A:

When the experimenter pushes Block A from behind, the block accelerates due to the force applied. This force is transmitted through the contact surface between Block A and Block B, causing Block B to also accelerate. The force exerted by the experimenter on Block A can be determined by analyzing the forces acting on both blocks.

Forces acting on Block A:
1. Force applied by the experimenter (F_experimenter): This is the force exerted by the experimenter on Block A to cause its acceleration.
2. Force exerted by Block A on Block B (F_AB): This force is a reaction force to the force applied by the experimenter. It is transmitted through the contact surface between the two blocks.

Forces acting on Block B:
1. Force exerted by Block B on Block A (F_BA): This is an equal and opposite reaction force to F_AB.
2. Normal force (N): This force arises due to the contact between the two blocks and acts perpendicular to the contact surface.
3. Weight (W_B): This force acts vertically downwards on Block B and is equal to the mass of Block B (mB) multiplied by the acceleration due to gravity (g).

Analysis:
Since the table is frictionless, there is no friction force acting between the blocks. Therefore, the only horizontal forces acting on the blocks are F_AB and F_BA.

By Newton's third law, F_AB is equal in magnitude but opposite in direction to F_BA. Thus, the force exerted by the experimenter on Block A (F_experimenter) is equal in magnitude but opposite in direction to the force exerted by Block A on Block B (F_AB).

Conclusion:
The force exerted by the experimenter on Block A is equal in magnitude but opposite in direction to the force exerted by Block A on Block B. This force is transmitted through the contact surface between the blocks and is the cause of the acceleration of both blocks.

Answer

(mA+mB)F/mB

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