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A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then maximum nd minimum value of force for which the block remains at rest?
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A block of mass 5 kg is placed on a horizontal surface with coefficien...
The maximum and minimum values of the force for which the block remains at rest are determined by the coefficient of friction and the mass of the block. The maximum force for which the block remains at rest is the force of static friction, which is given by the formula Fs = μsN, where Fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force. The normal force is equal to the weight of the block, which is the force of gravity acting on the block.
In this case, the maximum force for which the block remains at rest is Fs = 0.2 * 5 kg * 9.8 m/s2 = 9.8 N.
The minimum force for which the block remains at rest is the force of kinetic friction, which is given by the formula Fk = μkN, where Fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force. The coefficient of kinetic friction is typically slightly smaller than the coefficient of static friction.
In this case, we can assume that the coefficient of kinetic friction is 0.1, which is slightly smaller than the coefficient of static friction. Therefore, the minimum force for which the block remains at rest is Fk = 0.1 * 5 kg * 9.8 m/s2 = 4.9 N.
Therefore, the maximum and minimum values of the force for which the block remains at rest are 9.8 N and 4.9 N, respectively.
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A block of mass 5 kg is placed on a horizontal surface with coefficien...
Introduction:
The problem deals with determining the maximum and minimum force required to keep a block at rest on a horizontal surface with a coefficient of friction of 0.2.
Formula:
The maximum force required to keep the block at rest can be calculated using the formula Fmax = μN, where μ is the coefficient of friction, and N is the normal force acting on the block.
Explanation:
To keep the block at rest, the force applied to it must be equal and opposite to the force of friction acting on it. The force of friction is given by f = μN. If the applied force is less than the force of friction, the block will not move, and it will be at rest.
Calculation:
Given, the mass of the block is 5 kg, and the coefficient of friction is 0.2.
The normal force acting on the block is N = mg = 5 x 9.8 = 49 N.
The maximum force required to keep the block at rest is Fmax = μN = 0.2 x 49 = 9.8 N.
The minimum force required to keep the block at rest is zero.
Conclusion:
The maximum force required to keep the block at rest on a horizontal surface with a coefficient of friction of 0.2 is 9.8 N, and the minimum force required is zero.
The problem deals with determining the maximum and minimum force required to keep a block at rest on a horizontal surface with a coefficient of friction of 0.2.
Formula:
The maximum force required to keep the block at rest can be calculated using the formula Fmax = μN, where μ is the coefficient of friction, and N is the normal force acting on the block.
Explanation:
To keep the block at rest, the force applied to it must be equal and opposite to the force of friction acting on it. The force of friction is given by f = μN. If the applied force is less than the force of friction, the block will not move, and it will be at rest.
Calculation:
Given, the mass of the block is 5 kg, and the coefficient of friction is 0.2.
The normal force acting on the block is N = mg = 5 x 9.8 = 49 N.
The maximum force required to keep the block at rest is Fmax = μN = 0.2 x 49 = 9.8 N.
The minimum force required to keep the block at rest is zero.
Conclusion:
The maximum force required to keep the block at rest on a horizontal surface with a coefficient of friction of 0.2 is 9.8 N, and the minimum force required is zero.
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A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then maximum nd minimum value of force for which the block remains at rest?
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A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then maximum nd minimum value of force for which the block remains at rest? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then maximum nd minimum value of force for which the block remains at rest? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then maximum nd minimum value of force for which the block remains at rest?.
A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then maximum nd minimum value of force for which the block remains at rest? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then maximum nd minimum value of force for which the block remains at rest? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then maximum nd minimum value of force for which the block remains at rest?.
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