Question

A small block of mass $m$ is projected horizontally with speed $u$ where friction coefficient between block and plane is given by $\mu =cx$, where $x$ is displacement of the block on plane. Find maximum distance covered by the block

• A. $\frac{u}{\sqrt{cg}}$
• B. $\frac{u}{\sqrt{2cg}}$
• C. $\frac{2u}{\sqrt{cg}}$
• D. $\frac{u}{2\sqrt{cg}}$

Answer 1

The correct option is A $\frac{u}{\sqrt{cg}}$

Because $\mu =cx$

$u=u$

For mass 'm', N$=$mg

$f=\mu mg$

$a=\frac{\mu mg}{m}=\mu g$

Here $a=$ retardation by friction$=-\mu g=-cgx$

Because

$a=v\frac{dv}{dx}=-cgx$

${\int }_u^{o}vdv=-cg{\int }_o^{x}xdx$

$\Rightarrow {\left[\frac{v^2}{2}\right]}_u^o=-cg{\left(\frac{x^2}{2}\right)}_o^x$

$\frac{-u^2}{2}=-\frac{cg{x}^2}{2}$

$\Rightarrow x=\frac{u}{\sqrt{cg}}$.