A small block of mass m is projected horizont | vedclass

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Question

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=-

Vedclass · Accepted Answer

$\frac{u}{{\sqrt {cg} }}$

Vedclass · Answer

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

$a=v \frac{d v}{d x}=-c g x$

$\int_{u}^{o} v d v=-c g \int_{0}^{x} x d x$

$\Rightarrow\left[\frac{v^{2}}{2}\right]_{u}^{o}=-c g\left(\frac{x^{2}}{2}\right)_{o}^{x}$

$\frac{-u^{2}}{2}=-\frac{c g x^{2}}{2}$

$\Rightarrow x=\frac{u}{\sqrt{c g}}$

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

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