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A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction is
- a)0
- b)2 N
- c)1 N
- d)varies with the area of contact of the block
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A block of mass 1kg is placed on the floor. The coefficient of static ...
The applied force is less than the maximum value of static friction which will be (0.2)(1kg)(10) = 2N. So the force of friction is equal to the applied force.
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Most Upvoted Answer
A block of mass 1kg is placed on the floor. The coefficient of static ...
Given data:
Mass of the block, m = 1 kg
Coefficient of static friction, μ = 0.6
Acceleration of truck, a = 5 m/s²
Frictional force:
The frictional force acting on the block is given by the equation:
F_friction = μN
where N is the normal force acting on the block.
Normal force:
The normal force acting on the block is equal and opposite to the weight of the block. So, N = mg, where g is the acceleration due to gravity.
Substituting the values, we get:
N = mg = 1 kg × 9.8 m/s² = 9.8 N
Frictional force:
F_friction = μN = 0.6 × 9.8 N = 5.88 N
Since the acceleration of the truck is greater than the maximum static friction, the block will start to slide. So, the frictional force acting on the block will be the maximum static friction, which is equal to 5 N.
Therefore, option B is the correct answer.
Mass of the block, m = 1 kg
Coefficient of static friction, μ = 0.6
Acceleration of truck, a = 5 m/s²
Frictional force:
The frictional force acting on the block is given by the equation:
F_friction = μN
where N is the normal force acting on the block.
Normal force:
The normal force acting on the block is equal and opposite to the weight of the block. So, N = mg, where g is the acceleration due to gravity.
Substituting the values, we get:
N = mg = 1 kg × 9.8 m/s² = 9.8 N
Frictional force:
F_friction = μN = 0.6 × 9.8 N = 5.88 N
Since the acceleration of the truck is greater than the maximum static friction, the block will start to slide. So, the frictional force acting on the block will be the maximum static friction, which is equal to 5 N.
Therefore, option B is the correct answer.
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Community Answer
A block of mass 1kg is placed on the floor. The coefficient of static ...
The normal force on the box is 10 N therefore maximum static friction would be 10×0.2=2N.Static friction adjusts it's value according to the situation.Since the required force is 1n therefore answer is C
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A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction isa)0b)2 Nc)1 Nd)varies with the area of contact of the blockCorrect answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction isa)0b)2 Nc)1 Nd)varies with the area of contact of the blockCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction isa)0b)2 Nc)1 Nd)varies with the area of contact of the blockCorrect answer is option 'C'. Can you explain this answer?.
A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction isa)0b)2 Nc)1 Nd)varies with the area of contact of the blockCorrect answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction isa)0b)2 Nc)1 Nd)varies with the area of contact of the blockCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction isa)0b)2 Nc)1 Nd)varies with the area of contact of the blockCorrect answer is option 'C'. Can you explain this answer?.
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A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction isa)0b)2 Nc)1 Nd)varies with the area of contact of the blockCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction isa)0b)2 Nc)1 Nd)varies with the area of contact of the blockCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A block of mass 1kg is placed on the floor. The coefficient of static friction is 0.2. If a force of 1N is applied to the block, the force of friction isa)0b)2 Nc)1 Nd)varies with the area of contact of the blockCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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