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4-2.Friction
hard
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
$\frac{u}{{2\sqrt {cg} }}$
Solution
Because $\mu=c x$
$u=u$
For mass 'm', N=mg
$f=\mu m g$
$a=\frac{\mu m g}{m}=\mu g$
Here $a=$ retardation by friction $=-\mu g=-c g x$ Because
$a=v \frac{d v}{d x}=-c g x$
$\int_{u}^{o} v d v=-c g \int_{0}^{x} x d x$
$\Rightarrow\left[\frac{v^{2}}{2}\right]_{u}^{o}=-c g\left(\frac{x^{2}}{2}\right)_{o}^{x}$
$\frac{-u^{2}}{2}=-\frac{c g x^{2}}{2}$
$\Rightarrow x=\frac{u}{\sqrt{c g}}$
Standard 11
Physics
