Question

A block of mass $m$ is placed on a surface which is a locus of the equation $Y=2x^2$, where $Y$ is along the vertical direction and $x$ is along the horizontal direction. If coefficient of friction is 0.5, then maximum height above the ground at which the block can be placed without slipping is

• A. $\frac{1}{48}m$
• B. $\frac{1}{16}m$
• C. $\frac{1}{32}m$
• D. $\frac{1}{64}m$

Answer 1

The correct option is C $\frac{1}{32}m$

Given, $Y=2x^2,$ $\mu =0.5$
$\frac{dY}{dx}=4x$
As, slope = $\frac{dY}{dx}$ $=\tan \theta$
Limiting force; $\mu =\tan \theta$
So, $\frac{dY}{dx}=4x=0.5$
$x=\frac{1}{8}$
$Y=2\times \frac{1}{8}\times \frac{1}{8}=\frac{1}{32}m$