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When a body slides down from rest along a smooth inclined plane making an angle of $45^{\circ}$ with the horizontal, it takes time T. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time pT, where p is some number greater than 1. Calculate the coefficient of friction between the body and the rough plane.

Answers (1)

Since the body slides down from rest along a smooth plane viz. inclined at $45^{\circ}$ in time ‘T’.

u=0, s=s & t=T,

$a = g \sin 45^{\circ}=g/\sqrt{2}$

Now, $s = ut + \frac{1}{2} at^{2}$

$s=\frac{gT^{2}}{2\sqrt{2}}$

The motion of the body along a rough plane,

u=0,

$s=\frac{gT^{2}}{2\sqrt{2}}$

………… (i)

$ma= mg \sin45^{\circ} - f$

$= mg\frac{ 1}{\sqrt{2}} - \mu N$

= $\frac{mg}{\sqrt{2}} - \mu mg \cos 45^{\circ}$

$ma =\frac{ mg}{\sqrt{2}} [1-\mu ]$

$a = \frac{g}{\sqrt{2}} (1- \mu )$

Now,

$s = \frac{g}{2\sqrt{2}} (1-\mu ) p^{2}T^{2}$ …….. (ii)

It is given that in both cases the distance is equal

$\frac{gT^{2}}{2\sqrt{2}}=\frac{g}{2\sqrt{2}}\left ( 1- \mu \right )p^{2}T^{2}$

$1 = (1-\mu ) p^{2}$

$1 = p^{2} - \mu p^{2}$

$\mu p^{2} = p^{2} - 1$

$\mu =\left [ 1-\frac{1}{p^{2}} \right ]$

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infoexpert24

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