Question

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

• A. $m^0$
• B. m
• C. $m^2$
• D. $m^{-1}$

Answer 1

The correct option is A $m^0$

From work energy theorem we get,

$\frac{1}{2}m{v}^2=mgh-\mu mg\cos \theta \left(\frac{h}{\sin \theta }\right)$

Where $\left(\frac{h}{\sin \theta }\right)$ is the length of inclination, h=height of inclination and $\theta$ is the angle of inclination.

So simplifying the W.E.T we get

$v=\sqrt{2gh-2\mu g\cos \theta \left(\frac{h}{\sin \theta }\right)}=\sqrt{2gh-2\mu gh\left(\frac{\cos \theta }{\sin \theta }\right)}=\sqrt{2gh-2\mu gh\left(\cot \theta \right)}$

$v\alpha m^0$