Question

A body of mass m slides down a plane inclined at an angle $\alpha .$ The coefficient of friction is $\mu .$ Find the rate at which kinetic plus gravitational potential energy is dissipated at any time t is:

• A. $\mu m{g}^2\cos \alpha (2\sin \alpha -\mu \cos \alpha )t$
• B. $\mu m{g}^2\cos \alpha (2\sin \alpha -2\mu \cos \alpha )t$
• C. $\mu m{g}^2\cos \alpha (\sin \alpha -\mu \cos \alpha )t$
• D. $\mu m{g}^2\cos \alpha (\sin \alpha -2\mu \cos \alpha )t$

Answer 1

The correct option is C $\mu m{g}^2\cos \alpha (\sin \alpha -\mu \cos \alpha )t$

Forces acting on body are $mgsin\alpha$ downwards along the plane and $\mu mgcos\alpha$ upwards along the plane.

Thus $ma=mgsin\alpha -\mu mgcos\alpha$

$⟹a=\frac{dv}{dt}=g(sin\alpha -\mu cos\alpha )$

$⟹v=g(sin\alpha -\mu cos\alpha )t$

Rate of dissipation of kinetic energy plus potential energy= Rate of heat dissipation by action of friction

$=f_{r}v$

$=\mu mgcos\alpha \times v$

$=\mu m{g}^{2}cos\alpha (sin\alpha -\mu cos\alpha )t$