Question

A body of mass 2 kg is lying on a floor. The coefficient of static friction is 0.54. what will be the value of friction force if the force is 2.8N and $g = 10m{s^{ - 2}}$?
A. Zero
B. 2N
C. 2.8N
D. 8N

Answer 1

Hint: To solve this question firstly we have to find out the static friction force on the object. To find out friction force we should have some variables like mass of body, static friction coefficient. Mainly we will find normal force action on the body and multiply this normal force to static friction coefficient.

Complete step by step answer:
As we have a body of mass 2 kg
And static friction coefficient (${\mu _s}$) is 0.54.
Now we will draw that free body diagram of the body to find out friction force:

Now we can see in the figure that there is a force which is in downward direction. This is due to acceleration due to gravity of the body.
To balance the force due to acceleration due to gravity the normal force actually works on the body.
Hence normal force will be equal to mg to balance the body
$\therefore $ there is a horizontal force of 2.8 N on the body
So to balance this horizontal force friction force will be acted.
Now ${f_{\max }} = {\mu _s} \times N$
$\therefore N = mg$
Where ${f_{\max }}$ is the maximum friction force
${\mu _s}$ is the static friction coefficient
And N is the normal force
On putting all values:
$
  {f_{\max }} = 0.54 \times 2 \times 10 \\
  {f_{\max }} = 10.8N \\
 $
Now we have calculated the maximum friction force on the body.
$\therefore $friction force is much bigger than the horizontal force so, body will not move.
So, value of fiction force will be equal to net applied horizontal force
Hence, value of fiction force will be 2.8 N

Therefore, the correct answer will be option number C.

Note:
Friction force: friction force is defined as the net horizontal force which is applied by constituent particles of the body in the opposite direction to oppose the movement of the body by external applied force.
Value of friction force is the product of mass of the body and normal force action on the body.
Mathematically, ${f_{\max }} = {\mu _s} \times N$