Question: 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 :-
$\frac{1}{3} m$
$\frac{1}{2} \mathrm{~m}$
$\frac{1}{6} m$
$\frac{2}{3} m$
Correct Option: , 3
Solution:
