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Question

(A) The slope of the tangent to the curve at any point is given by $	an 	heta = \frac{dy}{dx}$.Given $y = \frac{x^3}{6}$,we have $\frac{dy}{dx} = \f

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$\frac{1}{6} 	ext{ m}$

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(A) The slope of the tangent to the curve at any point is given by $	an 	heta = \frac{dy}{dx}$.Given $y = \frac{x^3}{6}$,we have $\frac{dy}{dx} = \frac{3x^2}{6} = \frac{x^2}{2}$.Thus,$	an 	heta = \frac{x^2}{2}$.For the block to remain in equilibrium without slipping,the condition is $\mu \ge 	an 	heta$.The maximum height corresponds to the limiting case where $\mu = 	an 	heta$.Given $\mu = 0.5 = \frac{1}{2}$,we set $\frac{1}{2} = \frac{x^2}{2}$,which gives $x^2 = 1$,so $x = 1 	ext{ m}$.Substituting $x = 1$ into the equation for $y$,we get the maximum height:$y = \frac{(1)^3}{6} = \frac{1}{6} 	ext{ m}$.

$A$ block of mass $m$ is placed on a surface with a vertical cross-section given by $y = \frac{x^3}{6}$. If the coefficient of friction is $0.5$,the maximum height above the ground at which the block can be placed without slipping is

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