A body of mass $2$ $kg$ is placed on a horizontal surface having coefficient of kinetic friction $0.4$ and coefficient of static friction $0.5$. If a horizontal force of $2.5$ $N$ is applied on the body then the frictional force acting on the body will be ($g=10m{s}^{-2}$)

The correct option is D 2.5 N

Given:

$m=2kg{\mu }_k=.4{\mu }_s=.5F=2.5$

Normal force acting on the body

$N=mg$

$N=2\times 10$

$N=20N$

Static Frictional Force

$f_{max}=coefficientofstaticFriction\times Normalforce$

$f_{max}=0.5\times 20$

$f_{max}=10N$

As applied force is $2.5N$ which is less that static frictional force, hence the body will not move and the frictional force acting on it will be $=2.5N$