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A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then the maximum and minimum value of force f for which the block remains at rest are?
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A block of mass 5 kg is placed on a horizontal surface with coefficien...
Introduction

To determine the maximum and minimum force required for a block of mass 5 kg to remain at rest on a horizontal surface with a coefficient of friction of 0.2, we need to consider the concept of static friction and its relationship with the applied force.

Static Friction

Static friction is the force that opposes the relative motion or impending motion between two surfaces in contact with each other. It only comes into play when an external force is applied to an object. The maximum static friction force can be calculated using the formula:

Maximum Static Friction Force (Fmax) = μs * N

Where:
- μs is the coefficient of static friction
- N is the normal force exerted by the surface on the object (equal to the weight of the object in this case)

Normal Force

The normal force is the force exerted by a surface perpendicular to the contact surface. In this case, when the block is placed on a horizontal surface, the normal force exerted by the surface on the block is equal to its weight.

Normal Force (N) = mass * gravitational acceleration = 5 kg * 9.8 m/s² = 49 N

Maximum Force (Fmax)

To find the maximum force required for the block to remain at rest, we need to calculate the maximum static friction force using the formula mentioned earlier.

Maximum Static Friction Force (Fmax) = μs * N = 0.2 * 49 N = 9.8 N

Therefore, the maximum force (Fmax) that can be applied to the block without causing it to move is 9.8 N.

Minimum Force (Fmin)

The minimum force required to keep the block at rest is zero. This is because the block will remain at rest even without the application of any external force as long as the static friction force is greater than or equal to zero. Since the coefficient of friction is positive (0.2), the static friction force will always be greater than or equal to zero, allowing the block to remain at rest without any external force.

Therefore, the minimum force (Fmin) required to keep the block at rest is zero.

Conclusion

The maximum force (Fmax) that can be applied to the block without causing it to move is 9.8 N. On the other hand, the minimum force (Fmin) required to keep the block at rest 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 the maximum and minimum value of force f for which the block remains at rest are?
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A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2 then the maximum and minimum value of force f for which the block remains at rest are? 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 the maximum and minimum value of force f for which the block remains at rest are? 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 the maximum and minimum value of force f for which the block remains at rest are?.
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