A block of mass 10 kg rests on a horizontal floor. The acceleration due to gravity is 9.81 m/s². The coefficient of static friction between the floor and the block is 0.2. A horizontal force of 10 N is applied on the block as shown in the figure. The magnitude of force of friction (in N) on the block is _______.

The correct answer is 10.

Step-by-step Explanation:

1. Identify the given parameters from the problem and the diagram:
- Mass of the block, $m=10\text{kg}$
- Acceleration due to gravity, $g=9.81{\text{m/s}}^2$
- Coefficient of static friction, ${\mu }_s=0.2$
- Applied horizontal force, $F=10\text{N}$ (as shown pointing to the right in the diagram)

2. Calculate the normal reaction force:
The block is resting on a horizontal floor, so the normal force $N$ is equal to the weight of the block $W$:
$N=W=mg$
Substituting the given values:
$N=10\times 9.81=98.1\text{N}$

3. Determine the maximum static friction force (limiting friction):
The maximum friction force that can act before the block starts to move is given by:
$f_{s,\text{max}}={\mu }_{s}N$
Substituting the values:
$f_{s,\text{max}}=0.2\times 98.1=19.62\text{N}$

4. Analyze the state of motion and calculate the actual friction force:
- The applied horizontal force is $F=10\text{N}$.
- The maximum available static friction force is $f_{s,\text{max}}=19.62\text{N}$.
Since the applied force is less than the maximum static friction force ($F<f_{s,\text{max}}$), the block does not move and remains in static equilibrium.

Under static equilibrium, the actual force of static friction ($f_s$) exactly balances the applied force:
$f_s=F=10\text{N}$

Thus, the magnitude of the force of friction on the block is 10 N.