Blocks of mass 4, 8, and 24 kg are lined up from left to right in that order on a frictionless surface so each block is touching the next one. A rightward-pointing force of magnitude 15 N is applied to the left-most block.

a.)What is the magnitude of the force that the middle block exerts on the rightmost one?

b.)What is the magnitude of the force that the leftmost block exerts on the middle one?

c.)Suppose now that the left-right order of the blocks is reversed. Now find the magnitude of the force that the leftmost block exerts on the middle one?

To find the magnitude of the forces between the blocks, we need to consider Newton's laws of motion and the concept of inter-block forces.

a.) To determine the magnitude of the force that the middle block exerts on the rightmost one, we can use Newton's third law, which states that every action has an equal and opposite reaction.

1. The applied force of 15 N acts on the leftmost block.
2. Due to Newton's third law, the leftmost block exerts an equal and opposite force on the middle block.
3. As there are no external forces acting on the middle block, the rightmost block exerts the same magnitude of force on the middle block.

Therefore, the magnitude of the force that the middle block exerts on the rightmost block is 15 N.

b.) To find the magnitude of the force that the leftmost block exerts on the middle one, we can analyze the forces acting on the middle block.

1. The leftmost block applies a force on the middle block.
2. The middle block experiences a force from the leftmost block and a force from the rightmost block.
3. The net force on the middle block is zero because it is in equilibrium.
4. Therefore, the magnitude of the force that the leftmost block exerts on the middle one is equal to the magnitude of the force that the rightmost block exerts on the middle one, which is 15 N.

c.) When the left-right order of the blocks is reversed, the forces between the blocks will remain the same.

1. The applied force of 15 N acts on the rightmost block.
2. The rightmost block exerts an equal and opposite force on the middle block due to Newton's third law.
3. As there are no external forces acting on the middle block, the leftmost block exerts the same magnitude of force on the middle block as the applied force, which is 15 N.

Therefore, even when the block order is reversed, the magnitude of the force that the leftmost block exerts on the middle one remains at 15 N.

To solve this problem, we need to analyze the forces acting on each block and apply Newton's second law of motion (F = ma). Let's break down the problem step by step:

a) To find the magnitude of the force that the middle block exerts on the rightmost one, we'll use Newton's third law of motion. According to Newton's third law, the force exerted on an object by another object is equal in magnitude but opposite in direction to the force exerted on the second object by the first.

Since the leftmost block exerts a rightward-pointing force of 15 N on the middle block, the middle block exerts a leftward-pointing force of magnitude 15 N on the rightmost block.

Therefore, the magnitude of the force that the middle block exerts on the rightmost one is 15 N.

b) To find the magnitude of the force that the leftmost block exerts on the middle one, we'll again apply Newton's third law. The force exerted by the middle block on the leftmost block is equal in magnitude but opposite in direction to the force exerted by the leftmost block on the middle one.

Since the force applied to the leftmost block is 15 N, the magnitude of the force that the leftmost block exerts on the middle one is also 15 N.

c) When the left-right order of the blocks is reversed, the mass of each block remains the same but their positions change. This reversal of the order does not affect the magnitude of forces between the blocks.

Therefore, the magnitude of the force that the leftmost block exerts on the middle one remains the same, which is 15 N.

a) Keep applying F = ma. First you need the total acceleration, a. Get it by applying F = ma to all three blocks, using the externally applied F = 15 N

a = F/m = 15 N/36 kg = 0.4167 m/s^2
Then apply it again, for the rightmost block only.
F'' = 24 a = 10 N

b) F' = 32 a = 13.33 N

c) F = 12 a = 5 N