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A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is?
Most Upvoted Answer
A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of ...
**Problem Analysis:**
In this problem, we have a block A placed on top of a box B, which is resting on a table. We are given the masses of A and B, as well as the coefficients of friction between A and B, and between B and the table. We are also given that a horizontal force of 25 N is applied to box B. We need to find the friction force between A and B.
**Solution:**
To solve this problem, we can follow these steps:
**Step 1:**
First, let's calculate the normal force acting on block A. The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the weight of block A is acting downwards, so the normal force will act upwards and be equal in magnitude. The normal force can be calculated using the formula:
Normal force (N) = mass (m) * acceleration due to gravity (g)
Given that the mass of block A is 2 kg and the acceleration due to gravity is 9.8 m/s^2, we can calculate the normal force:
N = 2 kg * 9.8 m/s^2 = 19.6 N
**Step 2:**
Next, let's calculate the maximum friction force between block A and box B. The maximum friction force can be calculated using the formula:
Maximum friction force (Fmax) = coefficient of friction (μ) * normal force (N)
Given that the coefficient of friction between A and B is 0.2, we can calculate the maximum friction force:
Fmax = 0.2 * 19.6 N = 3.92 N
**Step 3:**
Now, let's consider the horizontal force applied to box B. We are given that a force of 25 N is applied horizontally to box B. This force will try to move block A along with box B.
**Step 4:**
Since the force applied is greater than the maximum friction force, block A will start moving along with box B. In this case, the friction force between A and B will be equal to the maximum friction force, which is 3.92 N.
Therefore, the friction force between block A and box B is 3.92 N.
In this problem, we have a block A placed on top of a box B, which is resting on a table. We are given the masses of A and B, as well as the coefficients of friction between A and B, and between B and the table. We are also given that a horizontal force of 25 N is applied to box B. We need to find the friction force between A and B.
**Solution:**
To solve this problem, we can follow these steps:
**Step 1:**
First, let's calculate the normal force acting on block A. The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the weight of block A is acting downwards, so the normal force will act upwards and be equal in magnitude. The normal force can be calculated using the formula:
Normal force (N) = mass (m) * acceleration due to gravity (g)
Given that the mass of block A is 2 kg and the acceleration due to gravity is 9.8 m/s^2, we can calculate the normal force:
N = 2 kg * 9.8 m/s^2 = 19.6 N
**Step 2:**
Next, let's calculate the maximum friction force between block A and box B. The maximum friction force can be calculated using the formula:
Maximum friction force (Fmax) = coefficient of friction (μ) * normal force (N)
Given that the coefficient of friction between A and B is 0.2, we can calculate the maximum friction force:
Fmax = 0.2 * 19.6 N = 3.92 N
**Step 3:**
Now, let's consider the horizontal force applied to box B. We are given that a force of 25 N is applied horizontally to box B. This force will try to move block A along with box B.
**Step 4:**
Since the force applied is greater than the maximum friction force, block A will start moving along with box B. In this case, the friction force between A and B will be equal to the maximum friction force, which is 3.92 N.
Therefore, the friction force between block A and box B is 3.92 N.
Community Answer
A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of ...
F= (u1+ u2) (m1+m2) g
here u is coefficient of friction
after solving we will get F=70N
here u is coefficient of friction
after solving we will get F=70N
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A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is?
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A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is? 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 A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is?.
A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is? 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 A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is?.
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A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is?, a detailed solution for A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is? has been provided alongside types of A 2kg block(A) is placed on 8 kg box (B) which rest on table coff. Of friction betwen A and B is0.2 and between table and B is 0.5 . A25 N horizontal force is applied on the block B ,then the friction force between A andB is? theory, EduRev gives you an
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