A block of mass $2\,\text{kg}$ rests on a rough inclined plane making an angle of $30^\circ$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$. The frictional force on the block is

Step 1: Identify the given parameters.

The physical quantities given for the block on the inclined plane are:

$$m = 2\text{ kg}$$ $$\theta = 30^\circ$$ $$\mu_s = 0.7$$ $$g = 9.8\text{ m/s}^2$$

Step 2: Calculate the component of gravity acting down the incline.

The force pulling the block down the slope is the parallel component of its weight.

$$F_{\parallel} = mg\sin\theta$$ $$F_{\parallel} = 2 \times 9.8 \times \sin(30^\circ)$$ $$F_{\parallel} = 9.8\text{ N}$$

Step 3: Calculate the maximum static frictional force.

First, determine the normal force acting on the block.

$$N = mg\cos\theta$$ $$N = 2 \times 9.8 \times \cos(30^\circ)$$ $$N \approx 16.97\text{ N}$$

Now, calculate the maximum limit of static friction.

$$f_{\max} = \mu_s N$$ $$f_{\max} = 0.7 \times 16.97$$ $$f_{\max} \approx 11.88\text{ N}$$

Step 4: Determine the actual frictional force acting on the block.

Compare the downward force along the incline with the maximum static friction.

$$F_{\parallel} < f_{\max}$$ $$9.8\text{ N} < 11.88\text{ N}$$

Since the force pulling the block down is less than the maximum static friction, the block does not slide and remains at rest. The static friction force self-adjusts to exactly balance the downward force.

$$f = F_{\parallel}$$ $$f = 9.8\text{ N}$$

Answer: The frictional force on the block is $9.8\text{ N}$ acting up the incline.