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

JIPMERJIPMER 2012Work, Energy and Power

Solution:

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From work energy principle,
$\frac{1}{2} mv^2 = mgh - f \times s$
$=mgh - (\mu mg \cos \theta ) \times \frac{h}{\sin \,\theta}$
$v = \sqrt{2gh - 2 \mu gh \cot \theta}$
Hence, $v \propto m^0$