A block of mass 200, g is kept stationary on | vedclass

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(A) The mass of the block is $m = 200\, g = 0.2\, kg$. The weight of the block is $W = mg = 0.2 	imes 10 = 2\, N$ (taking $g = 10\, m/s^2$).For the b

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$12$

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(A) The mass of the block is $m = 200\, g = 0.2\, kg$. The weight of the block is $W = mg = 0.2 	imes 10 = 2\, N$ (taking $g = 10\, m/s^2$).For the block to remain stationary on the smooth inclined plane,the forces acting along the plane must balance.The component of the weight acting down the plane is $mg \sin 60^{\circ}$.The component of the horizontal force $F$ acting up the plane is $F \cos 60^{\circ}$.Equating these forces for equilibrium: $F \cos 60^{\circ} = mg \sin 60^{\circ}$.Substituting the values: $F 	imes \frac{1}{2} = 2 	imes \frac{\sqrt{3}}{2}$.$F = 2\sqrt{3} = \sqrt{4 	imes 3} = \sqrt{12}$.Given $F = \sqrt{x}$,we have $\sqrt{x} = \sqrt{12}$,therefore $x = 12$.

$A$ 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 the figure. The value of $x = .....$

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