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:
JEE MAIN 2014, Difficult
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$ \tan \theta =\mu \frac{d y}{d x}=\frac{x^2}{2}(\text { from question })$
$ \text { coefficient of friction } \mu=0.5$
$ \therefore 0.5=\frac{x^2}{2}$
$ \Rightarrow x= \pm 1$
$ \text { Now }, y=\frac{x^3}{6}=\frac{1}{6} m$
$ \text { coefficient of friction } \mu=0.5$
$ \therefore 0.5=\frac{x^2}{2}$
$ \Rightarrow x= \pm 1$
$ \text { Now }, y=\frac{x^3}{6}=\frac{1}{6} m$
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