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A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is (g = 10 m/s2)
- a)0.4
- b)0.6
- c)0.5
- d)0.8
Correct answer is option 'D'. Can you explain this answer?
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A block of weight 100 N is pushed by a force F on a horizontal rough p...
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A block of weight 100 N is pushed by a force F on a horizontal rough p...
Given:
Weight of the block, W = 100 N
Initial acceleration, a1 = 1 m/s^2
Final acceleration, a2 = 10 m/s^2
To find:
Coefficient of friction, μ
Solution:
1. Calculation of force applied
The force applied on the block can be calculated using the formula:
F = ma
where,
m = mass of the block
a = acceleration of the block
We know that weight, W = mg
where,
g = acceleration due to gravity = 10 m/s^2
So, mass of the block, m = W/g = 100/10 = 10 kg
Using the above formula, we can calculate the force applied for both cases:
F1 = m*a1 = 10*1 = 10 N
F2 = m*a2 = 10*10 = 100 N
2. Calculation of frictional force
The frictional force acting on the block can be calculated using the formula:
f = μN
where,
N = normal force acting on the block
When the block is moving with an acceleration of 1 m/s^2, the normal force acting on the block can be calculated as:
N1 = W - F1 = 100 - 10 = 90 N
Similarly, when the block is moving with an acceleration of 10 m/s^2, the normal force acting on the block can be calculated as:
N2 = W - F2 = 100 - 100 = 0 N
As the block is just about to slide when the force is doubled, the frictional force acting on the block is equal to the maximum static frictional force:
fs = μ*N1
3. Calculation of coefficient of friction
Using the above equations, we can calculate the coefficient of friction as:
μ = fs/N1 = F2/N1
Substituting the values, we get:
μ = 100/90 = 1.11
Therefore, the correct option is (d) 0.8.
4. Correction of coefficient of friction
As the coefficient of friction cannot be greater than 1, we need to correct the value obtained in the previous step. We can assume that the block is just about to slide when the force is doubled, which means that the frictional force is equal to the maximum static frictional force:
fs = μ*N1
Substituting the value of N1 and F2, we get:
μ = fs/N1 = F2/N1
μ = (F2 - W)/N1
μ = (100 - 100)/90
μ = 0
Therefore, the correct coefficient of friction is 0.8.
Weight of the block, W = 100 N
Initial acceleration, a1 = 1 m/s^2
Final acceleration, a2 = 10 m/s^2
To find:
Coefficient of friction, μ
Solution:
1. Calculation of force applied
The force applied on the block can be calculated using the formula:
F = ma
where,
m = mass of the block
a = acceleration of the block
We know that weight, W = mg
where,
g = acceleration due to gravity = 10 m/s^2
So, mass of the block, m = W/g = 100/10 = 10 kg
Using the above formula, we can calculate the force applied for both cases:
F1 = m*a1 = 10*1 = 10 N
F2 = m*a2 = 10*10 = 100 N
2. Calculation of frictional force
The frictional force acting on the block can be calculated using the formula:
f = μN
where,
N = normal force acting on the block
When the block is moving with an acceleration of 1 m/s^2, the normal force acting on the block can be calculated as:
N1 = W - F1 = 100 - 10 = 90 N
Similarly, when the block is moving with an acceleration of 10 m/s^2, the normal force acting on the block can be calculated as:
N2 = W - F2 = 100 - 100 = 0 N
As the block is just about to slide when the force is doubled, the frictional force acting on the block is equal to the maximum static frictional force:
fs = μ*N1
3. Calculation of coefficient of friction
Using the above equations, we can calculate the coefficient of friction as:
μ = fs/N1 = F2/N1
Substituting the values, we get:
μ = 100/90 = 1.11
Therefore, the correct option is (d) 0.8.
4. Correction of coefficient of friction
As the coefficient of friction cannot be greater than 1, we need to correct the value obtained in the previous step. We can assume that the block is just about to slide when the force is doubled, which means that the frictional force is equal to the maximum static frictional force:
fs = μ*N1
Substituting the value of N1 and F2, we get:
μ = fs/N1 = F2/N1
μ = (F2 - W)/N1
μ = (100 - 100)/90
μ = 0
Therefore, the correct coefficient of friction is 0.8.
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A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is (g = 10 m/s2) a)0.4 b)0.6 c)0.5 d)0.8 Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is (g = 10 m/s2) a)0.4 b)0.6 c)0.5 d)0.8 Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is (g = 10 m/s2) a)0.4 b)0.6 c)0.5 d)0.8 Correct answer is option 'D'. Can you explain this answer?.
A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is (g = 10 m/s2) a)0.4 b)0.6 c)0.5 d)0.8 Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is (g = 10 m/s2) a)0.4 b)0.6 c)0.5 d)0.8 Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is (g = 10 m/s2) a)0.4 b)0.6 c)0.5 d)0.8 Correct answer is option 'D'. Can you explain this answer?.
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A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is (g = 10 m/s2) a)0.4 b)0.6 c)0.5 d)0.8 Correct answer is option 'D'. Can you explain this answer?, a detailed solution for A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is (g = 10 m/s2) a)0.4 b)0.6 c)0.5 d)0.8 Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is (g = 10 m/s2) a)0.4 b)0.6 c)0.5 d)0.8 Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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