question_answer

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

A) \[\frac{1}{2}m\]            

B) \[\frac{1}{6}m\]

C) \[\frac{2}{3}m\]            

D) \[\frac{1}{3}m\]

Correct Answer: B

Solution :

[b]: Block is under limiting friction, so \[\mu =\tan \theta \]                                           ?(i) Equation of the surface, \[y=\frac{{{x}^{3}}}{6}\] Slope,\[\frac{dy}{dx}=\frac{{{x}^{2}}}{2}\]                                        ?(ii) From eqns (i) and (ii),2 we get \[\mu =\frac{{{x}^{2}}}{2}\] Here, \[\mu =0.5\therefore 0.5=\frac{{{x}^{2}}}{2}\]or\[{{x}^{2}}=1\]or\[x=1\] So, \[y=\frac{1}{6}\]