d
\(\begin{array}{l}
\tan \theta = \mu \frac{{dy}}{{dx}} = \frac{{{x^2}}}{2}\left( {from\,question} \right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,coefficient\,of\,friction\,\mu = 0.5\\
\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,0.5 = \frac{{{x^2}}}{2}\\
\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,x = \pm 1\\
\,\,\,\,\,\,\,\,\,\,\,\,\,Now,\,y = \frac{{{x^3}}}{6} = \frac{1}{6}m
\end{array}\)
