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A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. The coefficient of friction is 0.6. The minimum force necessary to start the cube moving up the plane is
- a)5.4N
- b)10.8N
- c)2.7N
- d)None
Correct answer is option 'B'. Can you explain this answer?
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A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. ...
Given:
Weight of cube, W = 10N
Slope of inclined plane, θ = 3/5
Coefficient of friction, μ = 0.6
To find: Minimum force required to move the cube up the inclined plane
Assumptions:
1. The cube is at rest, so static friction acts on it.
2. The force required to move the cube up the plane is minimum force.
3. The entire weight of the cube acts vertically downwards.
Solution:
1. Resolving forces:
The weight of the cube can be resolved into two components:
- Force acting perpendicular to the plane, Wcosθ
- Force acting parallel to the plane, Wsinθ
The force due to friction acting on the cube will be μN, where N is the normal force acting perpendicular to the plane.
2. Calculating normal force:
The normal force acting on the cube can be calculated by resolving the weight of the cube perpendicular to the plane:
N = Wcosθ
N = 10N x 3/5
N = 6N
3. Calculating force due to friction:
The force due to friction acting on the cube can be calculated using the formula:
Force due to friction = μN
Force due to friction = 0.6 x 6N
Force due to friction = 3.6N
4. Calculating minimum force required to move the cube up the plane:
The minimum force required to move the cube up the plane can be calculated by resolving the weight of the cube parallel to the plane and adding the force due to friction acting on it:
Minimum force = Wsinθ + Force due to friction
Minimum force = 10N x 4/5 + 3.6N
Minimum force = 8N + 3.6N
Minimum force = 11.6N
Therefore, the minimum force required to move the cube up the inclined plane is 11.6N, which is closest to option (B), 10.8N.
Weight of cube, W = 10N
Slope of inclined plane, θ = 3/5
Coefficient of friction, μ = 0.6
To find: Minimum force required to move the cube up the inclined plane
Assumptions:
1. The cube is at rest, so static friction acts on it.
2. The force required to move the cube up the plane is minimum force.
3. The entire weight of the cube acts vertically downwards.
Solution:
1. Resolving forces:
The weight of the cube can be resolved into two components:
- Force acting perpendicular to the plane, Wcosθ
- Force acting parallel to the plane, Wsinθ
The force due to friction acting on the cube will be μN, where N is the normal force acting perpendicular to the plane.
2. Calculating normal force:
The normal force acting on the cube can be calculated by resolving the weight of the cube perpendicular to the plane:
N = Wcosθ
N = 10N x 3/5
N = 6N
3. Calculating force due to friction:
The force due to friction acting on the cube can be calculated using the formula:
Force due to friction = μN
Force due to friction = 0.6 x 6N
Force due to friction = 3.6N
4. Calculating minimum force required to move the cube up the plane:
The minimum force required to move the cube up the plane can be calculated by resolving the weight of the cube parallel to the plane and adding the force due to friction acting on it:
Minimum force = Wsinθ + Force due to friction
Minimum force = 10N x 4/5 + 3.6N
Minimum force = 8N + 3.6N
Minimum force = 11.6N
Therefore, the minimum force required to move the cube up the inclined plane is 11.6N, which is closest to option (B), 10.8N.
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A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. The coefficient of friction is 0.6. The minimum force necessary to start the cube moving up the plane isa)5.4Nb)10.8Nc)2.7Nd)NoneCorrect answer is option 'B'. Can you explain this answer?
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A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. The coefficient of friction is 0.6. The minimum force necessary to start the cube moving up the plane isa)5.4Nb)10.8Nc)2.7Nd)NoneCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. The coefficient of friction is 0.6. The minimum force necessary to start the cube moving up the plane isa)5.4Nb)10.8Nc)2.7Nd)NoneCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. The coefficient of friction is 0.6. The minimum force necessary to start the cube moving up the plane isa)5.4Nb)10.8Nc)2.7Nd)NoneCorrect answer is option 'B'. Can you explain this answer?.
A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. The coefficient of friction is 0.6. The minimum force necessary to start the cube moving up the plane isa)5.4Nb)10.8Nc)2.7Nd)NoneCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. The coefficient of friction is 0.6. The minimum force necessary to start the cube moving up the plane isa)5.4Nb)10.8Nc)2.7Nd)NoneCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cube of weight 10N rests on a rough inclined plane of slope 3 in 5. The coefficient of friction is 0.6. The minimum force necessary to start the cube moving up the plane isa)5.4Nb)10.8Nc)2.7Nd)NoneCorrect answer is option 'B'. Can you explain this answer?.
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