Solve: two blocks each of mass 3.0kg are connected by a light cord and placed on a rough horizontal surface.When a horizontal force of 20N is applied on a block,an acceleration of 0.50 m/s2 is produced in each block.Find the tension in the string and the frictional force which is same on the two blocks.?

Question

Answer

Is it frictional force =10N and T=8.5N using g=10m/s^2.use 20-(T+f)=3*0.5& f-T=3*0.5

Answer

Given:
- Mass of each block = 3.0 kg
- Applied force = 20 N
- Acceleration of each block = 0.50 m/s²

To Find:
- Tension in the string
- Frictional force on the blocks

Explanation:

1. Free Body Diagram:
Let's start by drawing the free body diagram for one of the blocks.

- There are three forces acting on each block:
- Tension force (T) in the string, pulling the blocks to the right.
- Applied force (20 N) to the right.
- Frictional force (F) opposing the motion to the left.
- The weight (mg) of each block acts downward, but it cancels out when considering the entire system since both blocks have the same weight.

2. Newton's Second Law:
According to Newton's second law of motion, the net force on an object is equal to the product of its mass and acceleration (F = ma).

- For each block, the net force can be written as:
- T - F = ma

3. Tension in the String:
Since the two blocks are connected by a light cord, the tension in the string is the same for both blocks. Let's consider the tension in the string as T.

- From the free body diagram, we can write the equation for the upper block:
- T - F = ma

- For the lower block, the equation becomes:
- T + F = ma

4. Solving the Equations:
We can solve the two equations simultaneously to find the values of T and F.

- Adding the two equations, we get:
- 2T = 2ma

- Substituting the given values:
- 2T = 2(3.0 kg)(0.50 m/s²)
- 2T = 3.0 N

- Solving for T:
- T = 1.5 N

5. Frictional Force:
To find the frictional force, we can substitute the value of T into one of the equations.

- Using the equation for the upper block:
- T - F = ma

- Substituting the values:
- 1.5 N - F = (3.0 kg)(0.50 m/s²)
- 1.5 N - F = 1.5 N

- Solving for F:
- F = 0 N

6. Conclusion:
Therefore, the tension in the string is 1.5 N, and the frictional force on the blocks is 0 N. The frictional force is zero because the applied force is exactly balanced by the tension in the string, resulting in no net force opposing the motion of the blocks.

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