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A block rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2)
- a)2.5
- b)4.0
- c)1.6
- d)2.0
Correct answer is option 'D'. Can you explain this answer?
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A block rests on a rough inclined plane making an angle of 30o with th...
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A block rests on a rough inclined plane making an angle of 30o with th...
Given:
Angle of inclined plane, θ = 30o
Coefficient of static friction, μs = 0.8
Frictional force, f = 10 N
Acceleration due to gravity, g = 10 m/s2
To find: Mass of the block, m
Assumptions:
The block is at rest and not moving down the inclined plane.
Solution:
1. Resolving forces:
The weight of the block acts vertically downwards and can be resolved into two components:
(i) mg sin θ which acts down the inclined plane
(ii) mg cos θ which acts perpendicular to the plane
The normal reaction, N, acts perpendicular to the plane and balances the component of weight mg cos θ.
The frictional force, f, acts up the inclined plane and opposes the motion of the block.
2. Applying equations of equilibrium:
The block is at rest. Therefore, the net force acting on it is zero. Hence, we can apply equations of equilibrium to solve for the mass of the block.
∑Fy = 0
N - mg cos θ = 0
N = mg cos θ
∑Fx = 0
f - mg sin θ = 0
f = mg sin θ
μs N = f
μs mg cos θ = mg sin θ
μs g cos θ = g sin θ
μs = tan θ
Substituting the given values, we get:
0.8 = tan 30o
Therefore, μs = tan 30o = 0.577
Substituting this value in the equation μs mg cos θ = mg sin θ, we get:
0.577 × m × g × cos 30o = m × g × sin 30o
0.5 × m × g = f
Substituting the value of f = 10 N, we get:
0.5 × m × 10 = 10
m = 2 kg
Therefore, the mass of the block is 2 kg.
Hence, the correct answer is option D.
Angle of inclined plane, θ = 30o
Coefficient of static friction, μs = 0.8
Frictional force, f = 10 N
Acceleration due to gravity, g = 10 m/s2
To find: Mass of the block, m
Assumptions:
The block is at rest and not moving down the inclined plane.
Solution:
1. Resolving forces:
The weight of the block acts vertically downwards and can be resolved into two components:
(i) mg sin θ which acts down the inclined plane
(ii) mg cos θ which acts perpendicular to the plane
The normal reaction, N, acts perpendicular to the plane and balances the component of weight mg cos θ.
The frictional force, f, acts up the inclined plane and opposes the motion of the block.
2. Applying equations of equilibrium:
The block is at rest. Therefore, the net force acting on it is zero. Hence, we can apply equations of equilibrium to solve for the mass of the block.
∑Fy = 0
N - mg cos θ = 0
N = mg cos θ
∑Fx = 0
f - mg sin θ = 0
f = mg sin θ
μs N = f
μs mg cos θ = mg sin θ
μs g cos θ = g sin θ
μs = tan θ
Substituting the given values, we get:
0.8 = tan 30o
Therefore, μs = tan 30o = 0.577
Substituting this value in the equation μs mg cos θ = mg sin θ, we get:
0.577 × m × g × cos 30o = m × g × sin 30o
0.5 × m × g = f
Substituting the value of f = 10 N, we get:
0.5 × m × 10 = 10
m = 2 kg
Therefore, the mass of the block is 2 kg.
Hence, the correct answer is option D.
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A block rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2) a)2.5 b)4.0 c)1.6 d)2.0 Correct answer is option 'D'. Can you explain this answer?
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A block rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2) a)2.5 b)4.0 c)1.6 d)2.0 Correct answer is option 'D'. 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 rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2) a)2.5 b)4.0 c)1.6 d)2.0 Correct answer is option 'D'. 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 rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2) a)2.5 b)4.0 c)1.6 d)2.0 Correct answer is option 'D'. Can you explain this answer?.
A block rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2) a)2.5 b)4.0 c)1.6 d)2.0 Correct answer is option 'D'. 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 rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2) a)2.5 b)4.0 c)1.6 d)2.0 Correct answer is option 'D'. 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 rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2) a)2.5 b)4.0 c)1.6 d)2.0 Correct answer is option 'D'. Can you explain this answer?.
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A block rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2) a)2.5 b)4.0 c)1.6 d)2.0 Correct answer is option 'D'. Can you explain this answer?, a detailed solution for A block rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2) a)2.5 b)4.0 c)1.6 d)2.0 Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of A block rests on a rough inclined plane making an angle of 30o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (Take g = 10 m/s2) a)2.5 b)4.0 c)1.6 d)2.0 Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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