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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A particle moves on a rough horizontal ground with some initial velocity $V_0$. If $\large\frac{3}{4}$th of its kinetic energy is lost in friction in time $t_0$ the coefficient of friction is

\[(a)\;\frac{v_0}{2gt_0} \quad (b)\;\frac{v_0}{4gt_0} \quad (c)\;\frac{3v_0}{4gt_0} \quad (d)\;\frac{v_0}{gt_0} \]

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1 Answer

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If energy becomes $\large\frac{1}{4}$ th .
$\large\frac{3}{4}$ th energy is lost, Velocity becomes $\large\frac{v_0}{2}$ under retardation $\mu g$
$\large\frac{v_0}{2}$$=v_0-\mu g t_0$
$\mu=\large\frac{v_0}{2gt_0}$
Hence a is the correct answer. 

 

answered Jul 15, 2013 by meena.p
edited Mar 21, 2014 by thagee.vedartham
 

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