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a 6.00-kg block is in contact with a 4.00-kg block on a horizontal frictionless surface as shown in the figure. the 6.00-kg block is being pushed by a horizontal 20.0-n force as shown. what is the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block?

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Answer:

Since the surface is frictionless, the only force acting on each block is the force of gravity, which we can ignore for now, and the force exerted by the other block.

We can use Newton's third law, which states that for every action, there is an equal and opposite reaction. Therefore, the force exerted by the 4.00-kg block on the 6.00-kg block is equal in magnitude and opposite in direction to the force exerted by the 6.00-kg block on the 4.00-kg block.

Now, let's focus on the 6.00-kg block. The force acting on it is the 20.0 N force to the right. Since the surface is frictionless, there is no opposing force, and the block accelerates to the right.

We can use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration. Therefore, we have:

Net force = mass x acceleration

20.0 N = 6.00 kg x acceleration

acceleration = 20.0 N / 6.00 kg = 3.33 m/s^2

Now, let's find the force exerted by the 6.00-kg block on the 4.00-kg block. We can use Newton's second law again, this time for the 4.00-kg block:

Net force = mass x acceleration

Force exerted by the 6.00-kg block on the 4.00-kg block = 4.00 kg x acceleration

Force exerted by the 6.00-kg block on the 4.00-kg block = 4.00 kg x 3.33 m/s^2

Force exerted by the 6.00-kg block on the 4.00-kg block = 13.3 N

Therefore, the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block is 13.3 N.

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