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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