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

• A. $\frac{1}{6}m$
• B. $\frac{2}{3}m$
• C. $\frac{1}{3}m$
• D. $\frac{1}{2}m$

Answer 1

The correct option is A $\frac{1}{6}m$

At equilibrium for limiting value of friction, $\mu =\tan \theta$

From the question,
$y=\frac{x^3}{6}$
$\Rightarrow \tan \theta =\frac{dy}{dx}=\frac{x^2}{2}$
$\therefore \mu =\frac{x^2}{2}$


Given $\mu =0.5$
$\Rightarrow 0.5=\frac{x^2}{2}$
$\Rightarrow x=1$

Therefore, maximum vertical height for no slipping $y=\frac{x^3}{6}=\frac{1}{6}m$