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(C) The slope of the curve at any point is given by $\frac{dy}{dx} = 	an 	heta$.Given $y = \frac{x^2}{\sqrt{3}}$,we have $\frac{dy}{dx} = \frac{2x}{

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$5$

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(C) The slope of the curve at any point is given by $\frac{dy}{dx} = 	an 	heta$.Given $y = \frac{x^2}{\sqrt{3}}$,we have $\frac{dy}{dx} = \frac{2x}{\sqrt{3}}$.At the release point,$y = \frac{1}{4\sqrt{3}}$,so $\frac{x^2}{\sqrt{3}} = \frac{1}{4\sqrt{3}}$,which gives $x^2 = \frac{1}{4}$,or $x = \frac{1}{2}$.Substituting $x = \frac{1}{2}$ into the slope equation: $	an 	heta = \frac{2(1/2)}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.Since $	an 	heta = \frac{1}{\sqrt{3}}$,we have $	heta = 30^\circ$.The acceleration of a block on a smooth inclined plane is $a = g \sin 	heta$.Thus,$a = 10 	imes \sin(30^\circ) = 10 	imes 0.5 = 5 \ m/s^2$.

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