# Help With Forces: Determine Force Exerted by 3.2kg Block

• potatosticks
In summary, the problem involves a 27-N force pushing a 6.5-kg block against a 3.2-kg block on a frictionless surface. The question is to determine the force that the 3.2-kg block exerts on the 6.5-kg block. After considering the 3.2-kg block's acceleration as 0, the conclusion is drawn that the 3.2-kg block exerts no force on the 6.5-kg block. However, it is clarified that the force of 27N affects both blocks as they slide on the table and they are in contact with each other. Further experimentation is needed to confirm the results.
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?

potatosticks said:
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.

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

## 1. How do I determine the force exerted by a 3.2kg block?

The force exerted by a 3.2kg block can be determined by using the formula F=ma, where F is the force, m is the mass of the block, and a is the acceleration. First, calculate the acceleration of the block by dividing the net force acting on it by its mass. Then, use this value of acceleration and the given mass of 3.2kg to calculate the force exerted by the block.

## 2. What is the unit of force used to measure the force exerted by a 3.2kg block?

The unit of force used to measure the force exerted by a 3.2kg block is Newtons (N). This unit is derived from the formula F=ma, where the unit of mass is kilograms (kg) and the unit of acceleration is meters per second squared (m/s^2).

## 3. How does the angle of the surface affect the force exerted by a 3.2kg block?

The angle of the surface does not directly affect the force exerted by a 3.2kg block. However, it can affect the normal force acting on the block, which is the force exerted by the surface on the block in a direction perpendicular to the surface. This normal force can then affect the net force and acceleration of the block, ultimately impacting the force exerted by the block.

## 4. Can the force exerted by a 3.2kg block change over time?

Yes, the force exerted by a 3.2kg block can change over time. This can happen if the net force acting on the block changes, causing its acceleration and therefore its force to change. The force exerted by the block can also change if the surface it is on or the angle of the surface changes, as explained in the previous answer.

## 5. How does the force exerted by a 3.2kg block compare to its weight?

The force exerted by a 3.2kg block can vary and may not always be equal to its weight. The weight of an object is the force of gravity acting on it, and it is calculated by multiplying the mass of the object by the acceleration due to gravity (9.8 m/s^2 on Earth). The force exerted by the block, on the other hand, is affected by other factors such as the net force and surface it is on, which can cause it to differ from its weight.

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