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 co-efficient of friction is $0.5$, the maximum height above the ground at which the block can be placed without slipping is

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

Answer 1

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

From the data given in the question,




Let $N$ be the normal force and $f$ is the friction force between block and given surface.

Here, Block is in static equilibrium, since it is not slipping.

From the given FBD,

$N=mg\cos \theta$ and $f=mg\sin \theta$

$\Rightarrow \mu N=mg\sin \theta \because f=\mu N$

$\Rightarrow \mu mg\cos \theta =mg\sin \theta$

$\therefore \mu =\tan \theta ............\left(1\right)$

Also,

Slope of the curve at the point where block is kept.

$m=\tan \theta =\frac{dy}{dx}=\frac{x^2}{2}$ ..........(2)

From Eq. $\left(1\right)$ and Eq. $\left(2\right)$

$\mu =\frac{x^2}{2}$

$\Rightarrow 0.5=\frac{x^2}{2}$

$\Rightarrow x=\pm 1\text{m}$

Substituting , $x=1\text{m}$ in $y=\frac{x^3}{6}$, we get,

$y=\frac{1}{6}\text{m}$

Hence, option (a) is the correct answer.