Answer
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Hint: The value of the static frictional force is given as the product of the coefficient of static friction and the normal force acting over the body. The body will be remaining at the rest itself as long as the force we apply is less than static frictional force. Therefore the maximum value of force applied up to which body is at rest will be the value of static frictional force. This information will help you in solving this question.
Complete step-by-step answer:
First of all the acceleration of the train up to which the static friction acts can be found using the equation,
$ma={{f}_{s}}\le {{\mu }_{s}}N={{\mu }_{s}}mg$
Where $m$ be the mass of the body, $a$ be the acceleration of the train, ${{f}_{s}}$ be the static frictional force, ${{\mu }_{s}}$ be coefficient of static friction, $N$ be the normal force acting on the body, $g$ be the acceleration due to gravity.
It has been mentioned in the question that the coefficient of static friction is given as,
${{\mu }_{s}}=0.2$
The body will be remaining at the rest itself as long as the force we apply is less than static frictional force.
$a\le {{\mu }_{s}}g$
Therefore the maximum value of force applied up to which body is at rest will be the value of static frictional force.
Therefore we can write that,
${{a}_{\max }}={{\mu }_{s}}g$
Substituting the values in it will give,
${{a}_{\max }}=0.2\times 10=2m{{s}^{-2}}$
Therefore the correct answer is mentioned as option A.
So, the correct answer is “Option A”.
Note: The coefficient of static friction is described as the ratio of the maximum static frictional force acting between the surfaces in contact just before the body moves to the normal force acting over it. This coefficient of friction is a dimensionless quantity.
Complete step-by-step answer:
First of all the acceleration of the train up to which the static friction acts can be found using the equation,
$ma={{f}_{s}}\le {{\mu }_{s}}N={{\mu }_{s}}mg$
Where $m$ be the mass of the body, $a$ be the acceleration of the train, ${{f}_{s}}$ be the static frictional force, ${{\mu }_{s}}$ be coefficient of static friction, $N$ be the normal force acting on the body, $g$ be the acceleration due to gravity.
It has been mentioned in the question that the coefficient of static friction is given as,
${{\mu }_{s}}=0.2$
The body will be remaining at the rest itself as long as the force we apply is less than static frictional force.
$a\le {{\mu }_{s}}g$
Therefore the maximum value of force applied up to which body is at rest will be the value of static frictional force.
Therefore we can write that,
${{a}_{\max }}={{\mu }_{s}}g$
Substituting the values in it will give,
${{a}_{\max }}=0.2\times 10=2m{{s}^{-2}}$
Therefore the correct answer is mentioned as option A.
So, the correct answer is “Option A”.
Note: The coefficient of static friction is described as the ratio of the maximum static frictional force acting between the surfaces in contact just before the body moves to the normal force acting over it. This coefficient of friction is a dimensionless quantity.