Help With Forces: Determine Force Exerted by 3.2kg Block

[potatosticks]

Ok the problem is:
A 27-N force pushes a 6.5-kg block against a 3.2-kg block. If the blocks are sliding on a horizontal, frictionless surface, determine the force that the 3.2-kg block exerts on the 6.5-kg block.

I'm thinking that the 3.2 block's acceleration equals 0, and so the force would have to equal zero. Does this mean the 3.2 block exerts no force on the 6.5 block?

 

[PhanthomJay]

Ok the problem is:
A 27-N force pushes a 6.5-kg block against a 3.2-kg block. If the blocks are sliding on a horizontal, frictionless surface, determine the force that the 3.2-kg block exerts on the 6.5-kg block.

I'm thinking that the 3.2 block's acceleration equals 0, and so the force would have to equal zero. Does this mean the 3.2 block exerts no force on the 6.5 block?

I think you are misreading the problem. The force of 27N pushes BOTH blocks along the table. The blocks are in contact with each other as they slide. Please try it again and report your results.

 

[kyle8921]

Based on the information provided, it is incorrect to assume that the 3.2-kg block exerts no force on the 6.5-kg block. In fact, the 3.2-kg block does exert a force on the 6.5-kg block, but it may not necessarily be equal to zero. This is because the two blocks are in contact with each other and are exerting a force on each other due to Newton's third law of motion. This force, known as the normal force, is responsible for keeping the blocks in contact and preventing them from passing through each other.

To determine the force exerted by the 3.2-kg block on the 6.5-kg block, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. In this case, the 6.5-kg block has an acceleration of 27 N / 6.5 kg = 4.15 m/s^2 due to the 27-N force pushing it. Since the two blocks are in contact and moving together, the 3.2-kg block will also have the same acceleration of 4.15 m/s^2. Using this acceleration and the mass of the 3.2-kg block, we can calculate the force exerted by the 3.2-kg block on the 6.5-kg block as follows:

Force = mass x acceleration
Force = 3.2 kg x 4.15 m/s^2
Force = 13.28 N

Therefore, the 3.2-kg block exerts a force of 13.28 N on the 6.5-kg block. It is important to note that this force is equal in magnitude but opposite in direction to the force exerted by the 6.5-kg block on the 3.2-kg block. This is in accordance with Newton's third law of motion.

In conclusion, it is incorrect to assume that the 3.2-kg block exerts no force on the 6.5-kg block. The two blocks are in contact with each other and are exerting equal and opposite forces on each other. By using Newton's second law of motion, we can calculate the force exerted by the 3.2-kg block on the 6.5-kg block, which is 13