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BITSAT 2020 Paper

Section: Physics

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Question : 34 of 150

Marks: +1, -0

A body of mass

2\text{ kg}

is placed on a horizontal surface having kinetic friction

0.4

and static friction 0.5. If the horizontal force applied on the body is

2.5\text{ N}

, the frictional force acting on the body will be

(g =10\text{ ms}^{2})

$8\text{ N}$
$10\text{ N}$
$20\text{ N}$
$2.5\text{ N}$

Solution:

The maximum static friction,

f_{s,\max} = \mu_{s} N =0.5 \times 2\text{ kg} \times 10\text{ ms}^{-2}

or,

f_{s,\max} =10\text{ N} >2.5\text{ N}

, which is greater than the applied horizontal force.

Hence, friction can impede the motion and the necessary required frictional force to balance the horizontal applied force will be equal.

Therefore,

f =2.5\text{ N}

.

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