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A block of mass 1kg slides down a rough inclined plane of inclination 60degree starting from its top. If coefficient of kinetic friction is 0.5 and length of the plane d = 1m then work done against friction is (1) 19.6 J (2) 2.45 J (3) 9.8 J (4) 4.9 J?
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A block of mass 1kg slides down a rough inclined plane of inclination ...
Given Data:
- Mass of the block, m = 1 kg
- Inclination of the plane, θ = 60°
- Coefficient of kinetic friction, μ = 0.5
- Length of the plane, d = 1 m
Step 1: Calculating the force of gravity along the inclined plane
The force of gravity acting on the block can be resolved into two components:
- Perpendicular to the inclined plane, mg⋅cos(θ)
- Parallel to the inclined plane, mg⋅sin(θ)
In this case, the force parallel to the inclined plane is responsible for the motion of the block.
Step 2: Calculating the force of friction
The force of friction can be determined using the equation:
F_friction = μ⋅N
Where N is the normal force acting on the block, which is equal to mg⋅cos(θ).
Therefore,
F_friction = μ⋅mg⋅cos(θ)
Step 3: Calculating the work done against friction
The work done against friction can be calculated using the equation:
W_friction = F_friction⋅d
Substituting the values,
W_friction = μ⋅mg⋅cos(θ)⋅d
Step 4: Calculating the numerical value
Substituting the given values into the equation:
W_friction = 0.5⋅1⋅9.8⋅cos(60°)⋅1
Using the values of cos(60°) and evaluating the expression:
W_friction = 0.5⋅1⋅9.8⋅0.5⋅1
W_friction = 2.45 J
Answer:
The work done against friction is 2.45 J. Hence, option (2) is the correct answer.
- Mass of the block, m = 1 kg
- Inclination of the plane, θ = 60°
- Coefficient of kinetic friction, μ = 0.5
- Length of the plane, d = 1 m
Step 1: Calculating the force of gravity along the inclined plane
The force of gravity acting on the block can be resolved into two components:
- Perpendicular to the inclined plane, mg⋅cos(θ)
- Parallel to the inclined plane, mg⋅sin(θ)
In this case, the force parallel to the inclined plane is responsible for the motion of the block.
Step 2: Calculating the force of friction
The force of friction can be determined using the equation:
F_friction = μ⋅N
Where N is the normal force acting on the block, which is equal to mg⋅cos(θ).
Therefore,
F_friction = μ⋅mg⋅cos(θ)
Step 3: Calculating the work done against friction
The work done against friction can be calculated using the equation:
W_friction = F_friction⋅d
Substituting the values,
W_friction = μ⋅mg⋅cos(θ)⋅d
Step 4: Calculating the numerical value
Substituting the given values into the equation:
W_friction = 0.5⋅1⋅9.8⋅cos(60°)⋅1
Using the values of cos(60°) and evaluating the expression:
W_friction = 0.5⋅1⋅9.8⋅0.5⋅1
W_friction = 2.45 J
Answer:
The work done against friction is 2.45 J. Hence, option (2) is the correct answer.
Community Answer
A block of mass 1kg slides down a rough inclined plane of inclination ...
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A block of mass 1kg slides down a rough inclined plane of inclination 60degree starting from its top. If coefficient of kinetic friction is 0.5 and length of the plane d = 1m then work done against friction is (1) 19.6 J (2) 2.45 J (3) 9.8 J (4) 4.9 J?
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A block of mass 1kg slides down a rough inclined plane of inclination 60degree starting from its top. If coefficient of kinetic friction is 0.5 and length of the plane d = 1m then work done against friction is (1) 19.6 J (2) 2.45 J (3) 9.8 J (4) 4.9 J? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A block of mass 1kg slides down a rough inclined plane of inclination 60degree starting from its top. If coefficient of kinetic friction is 0.5 and length of the plane d = 1m then work done against friction is (1) 19.6 J (2) 2.45 J (3) 9.8 J (4) 4.9 J? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 1kg slides down a rough inclined plane of inclination 60degree starting from its top. If coefficient of kinetic friction is 0.5 and length of the plane d = 1m then work done against friction is (1) 19.6 J (2) 2.45 J (3) 9.8 J (4) 4.9 J?.
A block of mass 1kg slides down a rough inclined plane of inclination 60degree starting from its top. If coefficient of kinetic friction is 0.5 and length of the plane d = 1m then work done against friction is (1) 19.6 J (2) 2.45 J (3) 9.8 J (4) 4.9 J? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A block of mass 1kg slides down a rough inclined plane of inclination 60degree starting from its top. If coefficient of kinetic friction is 0.5 and length of the plane d = 1m then work done against friction is (1) 19.6 J (2) 2.45 J (3) 9.8 J (4) 4.9 J? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 1kg slides down a rough inclined plane of inclination 60degree starting from its top. If coefficient of kinetic friction is 0.5 and length of the plane d = 1m then work done against friction is (1) 19.6 J (2) 2.45 J (3) 9.8 J (4) 4.9 J?.
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