A block of mass $m$ is placed on a surface with a vertical coss 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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$\tan \theta=\frac{d y}{d x}$
$\tan \theta=\frac{3 x^{2}}{6}$
$\tan \theta=\frac{1}{2} x^{2}$
at max height
$\mu m g \cos \theta=m g \sin \theta$
$\mu=\tan \theta$
$\frac{1}{2}=\frac{1}{2} x^{2}$
$x=\pm 1 m$
at $x=1$
$\mathrm{y}=\frac{1}{6} \mathrm{m}$
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