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A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is
- a)9.8 N
- b)0.7 × 9.8√3 N
- c)9.8 × √3 N
- d)0.7 × 9.8 N
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A block of mass 2 kg rests on a rough inclined plane making an angle o...
The force which is trying to accelerate the block on the inclined plane in the downward direction is mg sinθ, while the force opposing this is the force of friction f. The force of friction f can take any value for 0 to μmg cosθ. Here
As the dragging force is 9.8 N, the force of friction will grow up to this value to stop the block from sliding.
As the dragging force is 9.8 N, the force of friction will grow up to this value to stop the block from sliding.
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A block of mass 2 kg rests on a rough inclined plane making an angle o...
Degrees with the horizontal. The coefficient of friction between the block and the plane is 0.5. Determine the force of friction acting on the block and the net force acting on the block parallel to the incline.
To determine the force of friction acting on the block, we first need to calculate the normal force. The normal force is the force exerted by the inclined plane perpendicular to the surface of contact. It can be calculated using the equation:
Normal force = mass * gravitational acceleration * cos(angle)
Given that the mass of the block is 2 kg, the gravitational acceleration is approximately 9.8 m/s^2, and the angle is 30 degrees, we can calculate the normal force:
Normal force = 2 kg * 9.8 m/s^2 * cos(30 degrees)
Normal force = 2 kg * 9.8 m/s^2 * 0.866
Normal force = 16.97 N
The force of friction can be calculated using the equation:
Force of friction = coefficient of friction * normal force
Given that the coefficient of friction is 0.5 and the normal force is 16.97 N, we can calculate the force of friction:
Force of friction = 0.5 * 16.97 N
Force of friction = 8.485 N
The net force acting on the block parallel to the incline can be calculated using the equation:
Net force = (mass * gravitational acceleration) * sin(angle) - force of friction
Given that the mass is 2 kg, the gravitational acceleration is approximately 9.8 m/s^2, the angle is 30 degrees, and the force of friction is 8.485 N, we can calculate the net force:
Net force = (2 kg * 9.8 m/s^2) * sin(30 degrees) - 8.485 N
Net force = 19.6 N * 0.5 - 8.485 N
Net force = 9.8 N - 8.485 N
Net force = 1.315 N
Therefore, the force of friction acting on the block is 8.485 N and the net force acting on the block parallel to the incline is 1.315 N.
To determine the force of friction acting on the block, we first need to calculate the normal force. The normal force is the force exerted by the inclined plane perpendicular to the surface of contact. It can be calculated using the equation:
Normal force = mass * gravitational acceleration * cos(angle)
Given that the mass of the block is 2 kg, the gravitational acceleration is approximately 9.8 m/s^2, and the angle is 30 degrees, we can calculate the normal force:
Normal force = 2 kg * 9.8 m/s^2 * cos(30 degrees)
Normal force = 2 kg * 9.8 m/s^2 * 0.866
Normal force = 16.97 N
The force of friction can be calculated using the equation:
Force of friction = coefficient of friction * normal force
Given that the coefficient of friction is 0.5 and the normal force is 16.97 N, we can calculate the force of friction:
Force of friction = 0.5 * 16.97 N
Force of friction = 8.485 N
The net force acting on the block parallel to the incline can be calculated using the equation:
Net force = (mass * gravitational acceleration) * sin(angle) - force of friction
Given that the mass is 2 kg, the gravitational acceleration is approximately 9.8 m/s^2, the angle is 30 degrees, and the force of friction is 8.485 N, we can calculate the net force:
Net force = (2 kg * 9.8 m/s^2) * sin(30 degrees) - 8.485 N
Net force = 19.6 N * 0.5 - 8.485 N
Net force = 9.8 N - 8.485 N
Net force = 1.315 N
Therefore, the force of friction acting on the block is 8.485 N and the net force acting on the block parallel to the incline is 1.315 N.
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A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block isa)9.8 Nb)0.7 × 9.8√3 Nc)9.8 × √3 Nd)0.7 × 9.8 NCorrect answer is option 'A'. Can you explain this answer?
Question Description
A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block isa)9.8 Nb)0.7 × 9.8√3 Nc)9.8 × √3 Nd)0.7 × 9.8 NCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block isa)9.8 Nb)0.7 × 9.8√3 Nc)9.8 × √3 Nd)0.7 × 9.8 NCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block isa)9.8 Nb)0.7 × 9.8√3 Nc)9.8 × √3 Nd)0.7 × 9.8 NCorrect answer is option 'A'. Can you explain this answer?.
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