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(B) For the block to remain in equilibrium without slipping,the angle of inclination $	heta$ at the point of placement must satisfy the condition $

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

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(B) For the block to remain in equilibrium without slipping,the angle of inclination $	heta$ at the point of placement must satisfy the condition $	an 	heta \le \mu$.The maximum height corresponds to the limiting case where $	an 	heta = \mu = 0.5$.The slope of the surface at any point $(x, y)$ is given by $\frac{dy}{dx} = 	an 	heta$.Given the equation of the surface $y = \frac{x^3}{6}$,we find the slope:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^3}{6} ight) = \frac{3x^2}{6} = \frac{x^2}{2}$.Equating the slope to the coefficient of friction:$\frac{x^2}{2} = 0.5$$x^2 = 1$$x = 1$.Now,substitute $x = 1$ into the equation of the surface to find the maximum height $y$:$y = \frac{(1)^3}{6} = \frac{1}{6} \, m$.

$A$ 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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