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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.?
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Solve: two blocks each of mass 3.0kg are connected by a light cord and...
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
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Solve: two blocks each of mass 3.0kg are connected by a light cord and...
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.
- 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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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.?
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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.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 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.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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.?.
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.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 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.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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.?.
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