A small body was launched up an inclined plane set at an angle $$\alpha =15{}^\circ$$  against the horizontal. Find the coefficient of friction, if the time of the ascent of the body is  $$\eta =2.0$$ times less than the time of its descent.

Solution:

Case 1. When the body is lauched up:

Let k be the cofficeint of friction, u the velocity of projection and  $$\ell$$  the distance traversed along the incline. Retarding force on the block  $$=mg\sin \alpha +kmg\cos \alpha$$  and hence the retardation  $$=g\sin \alpha +kg\cos \alpha$$ .

Using the equation of  particle kinematics along the incline,  $$0={{u}^{2}}-2(g\sin \alpha +kg\cos \alpha )\ell$$

Or,  $$\ell =\frac{{{u}^{2}}}{2(g\sin \alpha +kg\cos \alpha )}\,\,\,\,\,\,\,\,\,(1)$$  and  $$0-u-(g\sin \alpha +kg\cos \alpha )t$$  or,  $$u=(g\sin \alpha +kg\cos \alpha )t\,\,\,\,(2)$$

Using (2) in (1)  $$\ell =\frac{1}{2}(g\sin \alpha +kg\cos \alpha ){{t}^{2}}\,\,\,\,\,\,\,(3)$$

Case (2). When the block comes downward, the net force on the body  $$=mg\sin \alpha -kmg\cos \alpha$$  and hence its acceleration  $$=g\sin \alpha -kg\cos \alpha$$ .

Let, t be the time required then,  $$\ell =\frac{1}{2}(g\sin \alpha -kg\cos \alpha ){{{t}’}^{2}}\,\,\,\,\,\,\,(4)$$

From Eqs. (3) and (4)  $$\frac{{{t}^{2}}}{{{{{t}’}}^{2}}}=\frac{\sin \alpha -k\cos \alpha }{\sin \alpha +k\cos \alpha }$$

But  $$\frac{t}{{{t}’}}=\frac{1}{\eta }$$ (according to the question),

Hence on solving we get  $$k=\frac{{{\eta }^{2}}-1}{{{\eta }^{2}}+1}\tan \alpha =0.16$$.

## Mechanics!

#### A point traversed half the distance with a velocity v0. The remaining part of the distance was covered with velocity v1 for half the time, and with velocity v2 for the other half of the time error: Content is protected !!