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An object of mass m is allowed to fall from rest along a rough inclined plane. The speed of the object at bottom of the plane is proportional to

\[(a)\;m^0 \quad (b)\;m \quad (c)\;m^2 \quad (d)\;m^{-1}\]

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Using work energy theorem
$\large\frac{1}{2}$$mv^2=mgh-(\mu mg \cos \theta)(\large\frac{h}{\sin \theta})$
where $\large\frac{h}{\sin \theta}$=length of the incline.
'h' hight of incline and $'\theta'$ angle of incline.
$v=\sqrt {2gh-2 \mu gh \cot \theta}$
$ v \;\alpha\; m^0$
Hence a is the correct answer. 

 

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