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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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An $8.0\;kg$ box is released on a $30^{\circ}$ inclined and accelerates down the incline at $30\;m/s^2$ Find the frictional force impending its motion . How large is the coefficient of friction in this situation ?

$\begin{array}{1 1}2.76\\0.54\\24 \\62\end{array} $

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Answer : 0.54
The component of the weight down the incline $=8(9.8) (0.5) =39.2 \;N$
Now $F=ma$ leads to $39.2 -f=8(0.3)$, So that $f= 36.8 \;N$ .
The Normal force equals the component of W perpendicular to the incline , $8(9.8)(0.867)=67.9\;N$.
Therefore $\mu =36.8/67.9 =0.54$
answered Aug 28, 2014 by meena.p
 

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