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A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, then
- a)Ms = 0.60
- b)Mk = 0.52
- c)Mk = 0.60
- d)Ms = 0.52
Correct answer is option 'A,B'. Can you explain this answer?
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Verified Answer
A block of mass 2.5 kg is kept on a rough horizontal surface. It is fo...
The weight of the mass of block, m = 2.5 kg
Now comparing the values of the weight of block = size of Normal force = mg = (2.5)(10) = 25 N
The horizontal force applied = 15 N (capable to move blocks)
{where d = 10 m, and t = 5 s, and "a" = block's acceleration}
The net force acting on the block of weight = ma = (2.5)(0.8) = 2.0 N
The force required to oppose the external force = 15 - 2 = 13 N
The value of the Kinetic Friction Force = 13 N
Therefore, dividing the Kinetic Force Friction and Horizontal Force by Normal Force we get
Coefficient of kinetic friction = 13/25 = 0.52 ANS
Coefficient of static friction = 15/25 = 0.60 ANS
Now comparing the values of the weight of block = size of Normal force = mg = (2.5)(10) = 25 N
The horizontal force applied = 15 N (capable to move blocks)
{where d = 10 m, and t = 5 s, and "a" = block's acceleration}
The net force acting on the block of weight = ma = (2.5)(0.8) = 2.0 N
The force required to oppose the external force = 15 - 2 = 13 N
The value of the Kinetic Friction Force = 13 N
Therefore, dividing the Kinetic Force Friction and Horizontal Force by Normal Force we get
Coefficient of kinetic friction = 13/25 = 0.52 ANS
Coefficient of static friction = 15/25 = 0.60 ANS
Most Upvoted Answer
A block of mass 2.5 kg is kept on a rough horizontal surface. It is fo...
To solve this problem, we need to analyze the forces acting on the block and apply Newton's laws of motion.
Given data:
- Mass of the block (m) = 2.5 kg
- Maximum horizontal force (Fmax) = 15 N
- Time taken (t) = 5 s
- Distance traveled (d) = 10 m
- Acceleration due to gravity (g) = 10 m/s^2
Analysis:
1. Determining the maximum static friction:
When the block is not sliding, the applied force is equal to the maximum static friction force. Therefore, we have:
Fmax = Fs
15 N = Fs
2. Determining the coefficient of static friction:
The static friction force can be determined using the equation:
Fs = μs * N
where μs is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the block, which is given by:
N = m * g
N = 2.5 kg * 10 m/s^2
N = 25 N
Substituting the values into the equation for Fs, we have:
15 N = μs * 25 N
Solving for μs, we get:
μs = 15 N / 25 N
μs = 0.6
Therefore, the coefficient of static friction (μs) is 0.6.
3. Determining the coefficient of kinetic friction:
When the block is sliding, the applied force is equal to the kinetic friction force. Therefore, we have:
F = Fk
15 N = Fk
The kinetic friction force can be determined using the equation:
Fk = μk * N
where μk is the coefficient of kinetic friction. Substituting the values into the equation, we have:
15 N = μk * 25 N
Solving for μk, we get:
μk = 15 N / 25 N
μk = 0.6
Therefore, the coefficient of kinetic friction (μk) is also 0.6.
4. Determining the acceleration:
The acceleration of the block can be determined using the equation:
a = (v - u) / t
where v is the final velocity, u is the initial velocity, and t is the time taken. In this case, the block starts from rest and travels a distance of 10 m. The final velocity can be determined using the equation:
v^2 = u^2 + 2as
where s is the distance traveled. Substituting the values into the equation, we have:
v^2 = 0 + 2a * 10 m
Simplifying the equation, we get:
v^2 = 20a m^2/s^2
Since the block starts from rest, the initial velocity (u) is 0 m/s. Therefore, the equation becomes:
v^2 = 20a m^2/s^2
The final velocity (v) can be determined using the equation:
v = u + at
Since the initial velocity (u) is 0 m/s, the equation becomes:
v = at
Substituting the values into the equation, we have:
v = a *
Given data:
- Mass of the block (m) = 2.5 kg
- Maximum horizontal force (Fmax) = 15 N
- Time taken (t) = 5 s
- Distance traveled (d) = 10 m
- Acceleration due to gravity (g) = 10 m/s^2
Analysis:
1. Determining the maximum static friction:
When the block is not sliding, the applied force is equal to the maximum static friction force. Therefore, we have:
Fmax = Fs
15 N = Fs
2. Determining the coefficient of static friction:
The static friction force can be determined using the equation:
Fs = μs * N
where μs is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the block, which is given by:
N = m * g
N = 2.5 kg * 10 m/s^2
N = 25 N
Substituting the values into the equation for Fs, we have:
15 N = μs * 25 N
Solving for μs, we get:
μs = 15 N / 25 N
μs = 0.6
Therefore, the coefficient of static friction (μs) is 0.6.
3. Determining the coefficient of kinetic friction:
When the block is sliding, the applied force is equal to the kinetic friction force. Therefore, we have:
F = Fk
15 N = Fk
The kinetic friction force can be determined using the equation:
Fk = μk * N
where μk is the coefficient of kinetic friction. Substituting the values into the equation, we have:
15 N = μk * 25 N
Solving for μk, we get:
μk = 15 N / 25 N
μk = 0.6
Therefore, the coefficient of kinetic friction (μk) is also 0.6.
4. Determining the acceleration:
The acceleration of the block can be determined using the equation:
a = (v - u) / t
where v is the final velocity, u is the initial velocity, and t is the time taken. In this case, the block starts from rest and travels a distance of 10 m. The final velocity can be determined using the equation:
v^2 = u^2 + 2as
where s is the distance traveled. Substituting the values into the equation, we have:
v^2 = 0 + 2a * 10 m
Simplifying the equation, we get:
v^2 = 20a m^2/s^2
Since the block starts from rest, the initial velocity (u) is 0 m/s. Therefore, the equation becomes:
v^2 = 20a m^2/s^2
The final velocity (v) can be determined using the equation:
v = u + at
Since the initial velocity (u) is 0 m/s, the equation becomes:
v = at
Substituting the values into the equation, we have:
v = a *
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A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. Can you explain this answer?
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A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. 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 of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. 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 of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. Can you explain this answer?.
A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. 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 of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. 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 of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. Can you explain this answer?.
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A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. Can you explain this answer?, a detailed solution for A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. Can you explain this answer? has been provided alongside types of A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is if found that it takes 5 second to slide throughout the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s2, thena)Ms= 0.60b)Mk= 0.52c)Mk= 0.60d)Ms= 0.52Correct answer is option 'A,B'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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