A block 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.

The correct option is D

2.5 N.

Explanation for the correct option

Step 1. Given data

Mass of block, $m=2kg$

Coefficient of kinetic friction, ${\mu }_k=0.4$

Coefficient of static friction, ${\mu }_s=0.5$

Force applied, $F=2.5N$

Step 2. Formula used

$F=\mu N$

$F=\mu mg$ ………….(a).

Step 3. Calculation of frictional force (f)

The free-body diagram of the given system of forces is shown below.

Using the corresponding values of the coefficient of limiting friction, mass and acceleration due to gravity in equation (a), we get.

Force of static friction, $F_s=0.5\times 2\times 10$

$F_s=10N$

clearly, $f<F_s$

But frictional force is a self-adjusting force and therefore, $f=F_s$.

$f=2.5N$

Hence, option (d) will be correct.