A block of mass $$200$$ g is kept stationary on a smooth inclined plane by applying a minimum horizontal force $$F = \sqrt{x}$$ N as shown in figure. The value of $$x$$ = ______.

Based on the provided image, we can accurately solve for x by balancing the forces acting on the block along the inclined plane.

1. Forces and Components

To keep the block stationary on a smooth incline, the net force along the plane must be zero.

• Component of Weight (mg): Acts down the incline as $$mg \sin(60^\circ)$$.
• Component of Horizontal Force (F): Acts up the incline as $$F \cos(60^\circ)$$.

2. Equilibrium Equation

For the block to remain stationary:

$$F \cos(60^\circ) = mg \sin(60^\circ)$$

$$F = mg \tan(60^\circ)$$

3. Numerical Calculation

Given:

• Mass (m) = $$200\text{ g} = 0.2\text{ kg}$$
• $$g = 10\text{ m/s}^2$$
• Weight (m
• g) = $$0.2 \times 10 = 2\text{ N}$$
• Angle ($$\theta$$) = $$60^\circ$$
• Force ($$F$$) = $$\sqrt{x}\text{ N}$$

Substituting the values:

$$\sqrt{x} = 2 \times \tan(60^\circ)$$

$$\sqrt{x} = 2 \times \sqrt{3}$$

Squaring both sides to find $x$:

$$x = (2\sqrt{3})^2$$$$x = 4 \times 3$$

$$\boxed{x = 12}$$