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A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge?
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A wedge of mass 10kg is placed over a smooth horizontal surface make a...
Given data:
- Mass of the wedge (m1) = 10 kg
- Angle of the wedge (θ) = 45°
- Mass of the block (m2) = 2 kg
- Coefficient of friction between the wedge and block (μ) = 0.6
- Applied force to prevent sliding downward (F1) = 30 N
- Applied force to prevent sliding upward (F2) = Maximum force
- Applied force to be analyzed (F3) = 20 N
Analysis:
1. To prevent the block from sliding downward, the force applied must be greater than or equal to the force of gravity acting on the block.
2. To prevent the block from sliding upward, the force applied must be greater than or equal to the force of gravity acting on the wedge and the block.
3. The force of gravity acting on an object is given by the equation: F_gravity = m * g, where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s^2).
Force analysis to prevent sliding downward:
- In this case, the force applied (F1) should be greater than or equal to the force of gravity acting on the block (m2 * g).
- The force of gravity acting on the block is given by: F_gravity2 = m2 * g = 2 kg * 9.8 m/s^2 = 19.6 N.
- Since F1 = 30 N is greater than F_gravity2, the block will not slide downward.
Force analysis to prevent sliding upward:
- In this case, the force applied (F2) should be greater than or equal to the force of gravity acting on the wedge and the block (m1 * g + m2 * g).
- The force of gravity acting on the wedge and the block is given by: F_gravity1 = m1 * g = 10 kg * 9.8 m/s^2 = 98 N, and F_gravity2 = m2 * g = 2 kg * 9.8 m/s^2 = 19.6 N.
- The total force of gravity acting on the system is: F_gravity_total = F_gravity1 + F_gravity2 = 98 N + 19.6 N = 117.6 N.
- Since F2 (maximum force) is greater than F_gravity_total, the block will not slide upward.
Force analysis for applied force of 20 N:
- In this case, the force applied (F3) is less than the force of gravity acting on the block (F_gravity2 = 19.6 N).
- Therefore, there will be a net force acting on the block, causing it to accelerate.
- The frictional force between the wedge and the block opposes the motion and can be calculated using the equation: F_friction = μ * F_normal, where μ is the coefficient of friction and F_normal is the normal force.
- The normal force can be calculated using the equation: F_normal = m2 * g * cos(θ), where m2 is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the wedge.
- Substituting the given
- Mass of the wedge (m1) = 10 kg
- Angle of the wedge (θ) = 45°
- Mass of the block (m2) = 2 kg
- Coefficient of friction between the wedge and block (μ) = 0.6
- Applied force to prevent sliding downward (F1) = 30 N
- Applied force to prevent sliding upward (F2) = Maximum force
- Applied force to be analyzed (F3) = 20 N
Analysis:
1. To prevent the block from sliding downward, the force applied must be greater than or equal to the force of gravity acting on the block.
2. To prevent the block from sliding upward, the force applied must be greater than or equal to the force of gravity acting on the wedge and the block.
3. The force of gravity acting on an object is given by the equation: F_gravity = m * g, where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s^2).
Force analysis to prevent sliding downward:
- In this case, the force applied (F1) should be greater than or equal to the force of gravity acting on the block (m2 * g).
- The force of gravity acting on the block is given by: F_gravity2 = m2 * g = 2 kg * 9.8 m/s^2 = 19.6 N.
- Since F1 = 30 N is greater than F_gravity2, the block will not slide downward.
Force analysis to prevent sliding upward:
- In this case, the force applied (F2) should be greater than or equal to the force of gravity acting on the wedge and the block (m1 * g + m2 * g).
- The force of gravity acting on the wedge and the block is given by: F_gravity1 = m1 * g = 10 kg * 9.8 m/s^2 = 98 N, and F_gravity2 = m2 * g = 2 kg * 9.8 m/s^2 = 19.6 N.
- The total force of gravity acting on the system is: F_gravity_total = F_gravity1 + F_gravity2 = 98 N + 19.6 N = 117.6 N.
- Since F2 (maximum force) is greater than F_gravity_total, the block will not slide upward.
Force analysis for applied force of 20 N:
- In this case, the force applied (F3) is less than the force of gravity acting on the block (F_gravity2 = 19.6 N).
- Therefore, there will be a net force acting on the block, causing it to accelerate.
- The frictional force between the wedge and the block opposes the motion and can be calculated using the equation: F_friction = μ * F_normal, where μ is the coefficient of friction and F_normal is the normal force.
- The normal force can be calculated using the equation: F_normal = m2 * g * cos(θ), where m2 is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the wedge.
- Substituting the given
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A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge?
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A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge?.
A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge?.
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A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge?, a detailed solution for A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge? has been provided alongside types of A wedge of mass 10kg is placed over a smooth horizontal surface make an angle 45^and a block of mass 2kg is placed over a wedge. If coefficient of friction between wedge and block is 0.6 and we applied minimum force 30N to not slide the block of mass 2kg downward and applied maximum force to not slide 2kg block upward now what if we would applied 20N force then acceleration of 2kg block with respect to 10kg wedge? theory, EduRev gives you an
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