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a small block of mass 1 kg is at rest on a parabolic surface given by y=x^2.the coefficient of friction between the block and surface is enough to prevent slipping .if position of block on surface is (2,4)then the normal force that the surface exerts at this position is?
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Calculating the Normal Force on the Surface:
Given Data:
- Mass of the block (m) = 1 kg
- Coefficient of friction (μ) = enough to prevent slipping
- Position of the block on the surface = (2, 4)
Analysis:
- The parabolic surface is given by y = x^2.
- At the given position (2, 4), the block is at rest.
Determination of Normal Force:
- The normal force exerted by the surface on the block is equal in magnitude and opposite in direction to the force of gravity acting on the block.
- The force of gravity acting on the block is given by Fg = m * g, where g is the acceleration due to gravity.
- At the position (2, 4), the gravitational force can be resolved into components perpendicular and parallel to the surface.
- The normal force can be calculated as N = m * g * cos(θ), where θ is the angle between the force of gravity and the normal force.
Solution:
- At position (2, 4), the slope of the surface is given by dy/dx = 2x.
- The angle between the force of gravity and the normal force is given by tan(θ) = dy/dx = 2x.
- Substituting x = 2 into the equation gives tan(θ) = 4.
- Therefore, θ = tan^(-1)(4).
- The normal force N = m * g * cos(θ) = 1 * 9.8 * cos(tan^(-1)(4)).
Final Answer:
- The normal force exerted by the surface at position (2, 4) is N = 1 * 9.8 * cos(tan^(-1)(4)).
Given Data:
- Mass of the block (m) = 1 kg
- Coefficient of friction (μ) = enough to prevent slipping
- Position of the block on the surface = (2, 4)
Analysis:
- The parabolic surface is given by y = x^2.
- At the given position (2, 4), the block is at rest.
Determination of Normal Force:
- The normal force exerted by the surface on the block is equal in magnitude and opposite in direction to the force of gravity acting on the block.
- The force of gravity acting on the block is given by Fg = m * g, where g is the acceleration due to gravity.
- At the position (2, 4), the gravitational force can be resolved into components perpendicular and parallel to the surface.
- The normal force can be calculated as N = m * g * cos(θ), where θ is the angle between the force of gravity and the normal force.
Solution:
- At position (2, 4), the slope of the surface is given by dy/dx = 2x.
- The angle between the force of gravity and the normal force is given by tan(θ) = dy/dx = 2x.
- Substituting x = 2 into the equation gives tan(θ) = 4.
- Therefore, θ = tan^(-1)(4).
- The normal force N = m * g * cos(θ) = 1 * 9.8 * cos(tan^(-1)(4)).
Final Answer:
- The normal force exerted by the surface at position (2, 4) is N = 1 * 9.8 * cos(tan^(-1)(4)).
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?a small block of mass 1 kg is at rest on a parabolic surface given by y=x^2.the coefficient of friction between the block and surface is enough to prevent slipping .if position of block on surface is (2,4)then the normal force that the surface exerts at this position is?
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?a small block of mass 1 kg is at rest on a parabolic surface given by y=x^2.the coefficient of friction between the block and surface is enough to prevent slipping .if position of block on surface is (2,4)then the normal force that the surface exerts at this position is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about ?a small block of mass 1 kg is at rest on a parabolic surface given by y=x^2.the coefficient of friction between the block and surface is enough to prevent slipping .if position of block on surface is (2,4)then the normal force that the surface exerts at this position is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?a small block of mass 1 kg is at rest on a parabolic surface given by y=x^2.the coefficient of friction between the block and surface is enough to prevent slipping .if position of block on surface is (2,4)then the normal force that the surface exerts at this position is?.
?a small block of mass 1 kg is at rest on a parabolic surface given by y=x^2.the coefficient of friction between the block and surface is enough to prevent slipping .if position of block on surface is (2,4)then the normal force that the surface exerts at this position is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about ?a small block of mass 1 kg is at rest on a parabolic surface given by y=x^2.the coefficient of friction between the block and surface is enough to prevent slipping .if position of block on surface is (2,4)then the normal force that the surface exerts at this position is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?a small block of mass 1 kg is at rest on a parabolic surface given by y=x^2.the coefficient of friction between the block and surface is enough to prevent slipping .if position of block on surface is (2,4)then the normal force that the surface exerts at this position is?.
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