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A body of mass 2 kg is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 N, then the frictional force acting on the body will be [g = 10 ms–2]
- a)8 N
- b)10 N
- c)20 N
- d)2.5 N
Correct answer is option 'D'. Can you explain this answer?
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A body of mass 2 kg is placed on a horizontal surface having kinetic f...
Limiting friction = 0.5 x 2 x 10= 10 N
The applied force is less than force of friction, therefore the force of friction is equal to the applied force.
The applied force is less than force of friction, therefore the force of friction is equal to the applied force.
Most Upvoted Answer
A body of mass 2 kg is placed on a horizontal surface having kinetic f...
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To find the frictional force acting on the body, we need to determine whether the body will experience static or kinetic friction. This depends on the force applied and the maximum frictional force that can be exerted by the surface.
The maximum static frictional force is given by:
$F_{s,max} = \mu_s N$
where $\mu_s$ is the static friction coefficient and $N$ is the normal force exerted on the body by the surface. The normal force is equal to the weight of the body, which is given by:
$N = mg$
where $m$ is the mass of the body and $g$ is the acceleration due to gravity.
Substituting the given values, we get:
$N = 2 \times 10 = 20$ N
$F_{s,max} = 0.5 \times 20 = 10$ N
Since the applied force of 2.5 N is less than the maximum static frictional force of 10 N, the body will experience static friction. Therefore, the frictional force acting on the body will be equal to the applied force, i.e.,
$F_f = 2.5$ N
To find the frictional force acting on the body, we need to determine whether the body will experience static or kinetic friction. This depends on the force applied and the maximum frictional force that can be exerted by the surface.
The maximum static frictional force is given by:
$F_{s,max} = \mu_s N$
where $\mu_s$ is the static friction coefficient and $N$ is the normal force exerted on the body by the surface. The normal force is equal to the weight of the body, which is given by:
$N = mg$
where $m$ is the mass of the body and $g$ is the acceleration due to gravity.
Substituting the given values, we get:
$N = 2 \times 10 = 20$ N
$F_{s,max} = 0.5 \times 20 = 10$ N
Since the applied force of 2.5 N is less than the maximum static frictional force of 10 N, the body will experience static friction. Therefore, the frictional force acting on the body will be equal to the applied force, i.e.,
$F_f = 2.5$ N
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A body of mass 2 kg is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 N, then the frictional force acting on the body will be [g = 10 ms–2]a)8 Nb)10 Nc)20 Nd)2.5 NCorrect answer is option 'D'. Can you explain this answer?
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A body of mass 2 kg is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 N, then the frictional force acting on the body will be [g = 10 ms–2]a)8 Nb)10 Nc)20 Nd)2.5 NCorrect answer is option 'D'. 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 body of mass 2 kg is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 N, then the frictional force acting on the body will be [g = 10 ms–2]a)8 Nb)10 Nc)20 Nd)2.5 NCorrect answer is option 'D'. 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 body of mass 2 kg is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 N, then the frictional force acting on the body will be [g = 10 ms–2]a)8 Nb)10 Nc)20 Nd)2.5 NCorrect answer is option 'D'. Can you explain this answer?.
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