A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

The correct option is 9.8 N.

Step-by-step Explanation:

1. Identify the given values:
Mass of the block, $m=2\text{kg}$
Angle of inclination, $\theta ={30}^{\circ }$
Coefficient of static friction, ${\mu }_s=0.7$
Acceleration due to gravity, $g=9.8{\text{m/s}}^2$

2. Determine the force components acting on the block:
The force acting down the inclined plane tending to slide the block is the component of gravitational force parallel to the incline:
$F_{\text{down}}=mg\sin \left(\theta \right)$

Substituting the given values:
$F_{\text{down}}=2\times 9.8\times \sin \left({30}^{\circ }\right)$
Since $\sin \left({30}^{\circ }\right)=0.5$:
$F_{\text{down}}=2\times 9.8\times 0.5=9.8\text{N}$

3. Calculate the maximum static frictional force (limiting friction):
The normal reaction force $N$ perpendicular to the incline is:
$N=mg\cos \left(\theta \right)$

The maximum force of static friction that can act to prevent motion is:
$f_{\text{max}}={\mu }_{s}N={\mu }_{s}mg\cos \left(\theta \right)$

Substituting the values:
$f_{\text{max}}=0.7\times 2\times 9.8\times \cos \left({30}^{\circ }\right)$
$f_{\text{max}}=0.7\times 2\times 9.8\times \frac{\sqrt{3}}{2}\approx 11.88\text{N}$

4. Analyze the state of the block:
Static friction is self-adjusting. It only acts with as much force as is required to prevent the object from moving, up to its maximum value $f_{\text{max}}$.

Since the downward force pulling the block ($F_{\text{down}}=9.8\text{N}$) is less than the maximum possible static frictional force ($f_{\text{max}}\approx 11.88\text{N}$), the block remains at rest (in equilibrium).

Therefore, the actual frictional force acting on the block is exactly equal to the force tending to pull it down the incline to maintain equilibrium:
$f=F_{\text{down}}=9.8\text{N}$