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Two Block A and B of masses 5 kg and 3kg respectively rest on a smooth horizontal surface which B over A the coefficient of friction between A and B is 0.5 the maximum horizontal force that can be applied to a so that there will be motion of Aand B without relative slipping is?
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Problem Statement:
Two blocks, A and B, with masses 5 kg and 3 kg respectively, rest on a smooth horizontal surface. Block B is placed on top of block A. The coefficient of friction between block A and B is 0.5. We need to determine the maximum horizontal force that can be applied to block A without causing relative slipping between the blocks.

Solution:

1. Free Body Diagrams:
To solve this problem, let's start by drawing the free body diagrams for both blocks A and B.

For block A:
- There is a gravitational force acting downwards (mg).
- There is a normal force exerted by block B (N).
- There is a frictional force between block A and B (f).

For block B:
- There is a gravitational force acting downwards (mg).
- There is a normal force exerted by the surface (N).

2. Equations of Motion:
Now, let's analyze the forces acting on block A.

Forces in the vertical direction:
- N - mg = 0 (Since block A is not moving vertically)

Forces in the horizontal direction:
- f - F = ma (Newton's second law, where F is the applied force and a is the acceleration of block A)

For block B, since it is not moving horizontally, the sum of forces in the horizontal direction is zero.

3. Frictional Force:
The frictional force (f) between block A and B can be calculated using the equation:

f = μN

Where μ is the coefficient of friction and N is the normal force. Since the normal force (N) is equal to the weight of block B (mg), we can rewrite the equation as:

f = μmg

4. Maximum Applied Force:
To find the maximum force (F) that can be applied to block A without causing relative slipping between the blocks, we need to consider the maximum frictional force that can be exerted between the blocks.

The maximum frictional force (f_max) is given by:

f_max = μN = μmg

Therefore, the maximum force (F) that can be applied to block A is equal to the maximum frictional force:

F = f_max = μmg

Substituting the given values, we have:

F = 0.5 * 5 kg * 9.8 m/s^2

Simplifying the equation, we find:

F = 24.5 N

Therefore, the maximum horizontal force that can be applied to block A without causing relative slipping between the blocks is 24.5 Newtons.
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Two Block A and B of masses 5 kg and 3kg respectively rest on a smooth horizontal surface which B over A the coefficient of friction between A and B is 0.5 the maximum horizontal force that can be applied to a so that there will be motion of Aand B without relative slipping is?
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Two Block A and B of masses 5 kg and 3kg respectively rest on a smooth horizontal surface which B over A the coefficient of friction between A and B is 0.5 the maximum horizontal force that can be applied to a so that there will be motion of Aand B without relative slipping 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 Two Block A and B of masses 5 kg and 3kg respectively rest on a smooth horizontal surface which B over A the coefficient of friction between A and B is 0.5 the maximum horizontal force that can be applied to a so that there will be motion of Aand B without relative slipping is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two Block A and B of masses 5 kg and 3kg respectively rest on a smooth horizontal surface which B over A the coefficient of friction between A and B is 0.5 the maximum horizontal force that can be applied to a so that there will be motion of Aand B without relative slipping is?.
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