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A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4 respectively. On applying the same force, what is the acceleration :
- a)0.98 m/s2
- b)9.8 m/s2
- c)0.54 m/s2
- d)5.292 m/s2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A body of mass 60 kg is dragged with just enough force to start moving...
Correct option (a)
Explanation:
Explanation:
Most Upvoted Answer
A body of mass 60 kg is dragged with just enough force to start moving...
Given:
Mass of the body (m) = 60 kg
Coefficient of static friction (μs) = 0.5
Coefficient of kinetic friction (μk) = 0.4
To find: Acceleration of the body on applying the same force.
Explanation:
When a force is applied to the body, there are two possible cases to consider: static friction and kinetic friction.
1. Static Friction:
Static friction acts when the body is at rest and the applied force is not enough to overcome the frictional force. The maximum value of static friction is given by:
F_static_max = μs * N
where N is the normal force.
In this case, the body is dragged with just enough force to start moving. So the applied force is equal to the maximum static friction force:
F_applied = F_static_max
2. Kinetic Friction:
Once the body starts moving, the frictional force changes to kinetic friction. The value of kinetic friction is given by:
F_kinetic = μk * N
Applying Newton's Second Law of Motion:
The net force acting on the body is given by:
F_net = F_applied - F_friction
When the body is at rest, F_applied = F_friction (maximum static friction).
So, F_net = F_applied - F_static_max = 0
When the body is in motion, F_applied > F_friction (kinetic friction).
So, F_net = F_applied - F_kinetic
The acceleration of the body is given by:
F_net = m * a
Solving the equations:
When the body is at rest (static friction):
F_applied - F_static_max = 0
F_applied = F_static_max = μs * N
When the body is in motion (kinetic friction):
F_applied - F_kinetic = m * a
F_applied = F_kinetic + m * a
μs * N = μk * N + m * a
Using the equation μs * N = μk * N + m * a, we can rearrange and solve for acceleration (a):
a = (μs * N - μk * N) / m
a = (μs - μk) * (N / m)
Since N is equal to the weight of the body (mg):
a = (μs - μk) * (mg / m)
a = (μs - μk) * g
Substituting the given values:
a = (0.5 - 0.4) * 9.8
a = 0.1 * 9.8
a = 0.98 m/s²
Therefore, the acceleration of the body when the same force is applied is 0.98 m/s². Thus, the correct answer is option 'A'.
Mass of the body (m) = 60 kg
Coefficient of static friction (μs) = 0.5
Coefficient of kinetic friction (μk) = 0.4
To find: Acceleration of the body on applying the same force.
Explanation:
When a force is applied to the body, there are two possible cases to consider: static friction and kinetic friction.
1. Static Friction:
Static friction acts when the body is at rest and the applied force is not enough to overcome the frictional force. The maximum value of static friction is given by:
F_static_max = μs * N
where N is the normal force.
In this case, the body is dragged with just enough force to start moving. So the applied force is equal to the maximum static friction force:
F_applied = F_static_max
2. Kinetic Friction:
Once the body starts moving, the frictional force changes to kinetic friction. The value of kinetic friction is given by:
F_kinetic = μk * N
Applying Newton's Second Law of Motion:
The net force acting on the body is given by:
F_net = F_applied - F_friction
When the body is at rest, F_applied = F_friction (maximum static friction).
So, F_net = F_applied - F_static_max = 0
When the body is in motion, F_applied > F_friction (kinetic friction).
So, F_net = F_applied - F_kinetic
The acceleration of the body is given by:
F_net = m * a
Solving the equations:
When the body is at rest (static friction):
F_applied - F_static_max = 0
F_applied = F_static_max = μs * N
When the body is in motion (kinetic friction):
F_applied - F_kinetic = m * a
F_applied = F_kinetic + m * a
μs * N = μk * N + m * a
Using the equation μs * N = μk * N + m * a, we can rearrange and solve for acceleration (a):
a = (μs * N - μk * N) / m
a = (μs - μk) * (N / m)
Since N is equal to the weight of the body (mg):
a = (μs - μk) * (mg / m)
a = (μs - μk) * g
Substituting the given values:
a = (0.5 - 0.4) * 9.8
a = 0.1 * 9.8
a = 0.98 m/s²
Therefore, the acceleration of the body when the same force is applied is 0.98 m/s². Thus, the correct answer is option 'A'.
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A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4 respectively. On applying the same force, what is the acceleration :a)0.98 m/s2b)9.8 m/s2c)0.54 m/s2d)5.292 m/s2Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4 respectively. On applying the same force, what is the acceleration :a)0.98 m/s2b)9.8 m/s2c)0.54 m/s2d)5.292 m/s2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4 respectively. On applying the same force, what is the acceleration :a)0.98 m/s2b)9.8 m/s2c)0.54 m/s2d)5.292 m/s2Correct answer is option 'A'. Can you explain this answer?.
A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4 respectively. On applying the same force, what is the acceleration :a)0.98 m/s2b)9.8 m/s2c)0.54 m/s2d)5.292 m/s2Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4 respectively. On applying the same force, what is the acceleration :a)0.98 m/s2b)9.8 m/s2c)0.54 m/s2d)5.292 m/s2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4 respectively. On applying the same force, what is the acceleration :a)0.98 m/s2b)9.8 m/s2c)0.54 m/s2d)5.292 m/s2Correct answer is option 'A'. Can you explain this answer?.
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