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A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is μ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10μ, then N is
[jee 2011]
Correct answer is '5N'. Can you explain this answer?
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A block is moving on an inclined plane making an angle 45º with h...
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A block is moving on an inclined plane making an angle 45º with h...
Degrees with the horizontal. The block has a weight of 10 N and is subjected to a frictional force of 4 N opposing its motion. To find the block's acceleration, we can break down the forces acting on it and use Newton's second law of motion.
First, let's resolve the weight of the block into components parallel and perpendicular to the inclined plane. The weight acting perpendicular to the plane is equal to the normal force, which is given by:
Normal force = Weight * cos(angle) = 10 N * cos(45°) = 10 N * 0.707 = 7.07 N
Next, let's resolve the weight parallel to the inclined plane. This component of the weight will act down the incline and will be opposed by the frictional force. The component of the weight acting down the incline is:
Weight parallel to the plane = Weight * sin(angle) = 10 N * sin(45°) = 10 N * 0.707 = 7.07 N
Now, let's calculate the net force acting on the block parallel to the inclined plane:
Net force parallel to the plane = Weight parallel to the plane - Frictional force
= 7.07 N - 4 N = 3.07 N
Finally, we can use Newton's second law of motion to find the acceleration of the block:
Net force parallel to the plane = mass * acceleration
3.07 N = mass * acceleration
Since the mass of the block is not given, we cannot determine the exact value of the acceleration without additional information.
First, let's resolve the weight of the block into components parallel and perpendicular to the inclined plane. The weight acting perpendicular to the plane is equal to the normal force, which is given by:
Normal force = Weight * cos(angle) = 10 N * cos(45°) = 10 N * 0.707 = 7.07 N
Next, let's resolve the weight parallel to the inclined plane. This component of the weight will act down the incline and will be opposed by the frictional force. The component of the weight acting down the incline is:
Weight parallel to the plane = Weight * sin(angle) = 10 N * sin(45°) = 10 N * 0.707 = 7.07 N
Now, let's calculate the net force acting on the block parallel to the inclined plane:
Net force parallel to the plane = Weight parallel to the plane - Frictional force
= 7.07 N - 4 N = 3.07 N
Finally, we can use Newton's second law of motion to find the acceleration of the block:
Net force parallel to the plane = mass * acceleration
3.07 N = mass * acceleration
Since the mass of the block is not given, we cannot determine the exact value of the acceleration without additional information.
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A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is μ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10μ, then N is[jee 2011]Correct answer is '5N'. Can you explain this answer?
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A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is μ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10μ, then N is[jee 2011]Correct answer is '5N'. Can you explain this answer? 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 A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is μ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10μ, then N is[jee 2011]Correct answer is '5N'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is μ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10μ, then N is[jee 2011]Correct answer is '5N'. Can you explain this answer?.
A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is μ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10μ, then N is[jee 2011]Correct answer is '5N'. Can you explain this answer? 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 A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is μ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10μ, then N is[jee 2011]Correct answer is '5N'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block is moving on an inclined plane making an angle 45º with horizontal and the coefficient of friciton is μ. the force required to just push it up the inclined plane is 3 times the force requried to just prevent it from sliding down. If we define N = 10μ, then N is[jee 2011]Correct answer is '5N'. Can you explain this answer?.
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