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A given object takes n times more time to slide down a $45^0$ rough inclined plane as it takes to slide down a perfectly smooth $45^0$ incline. The coefficient of kinetic friction between the object and the incline is :

Option 1)
$\frac{1 }{2-n^2}$

Option 2)
$1-\frac{1}{n^{2}}$

Option 3)
$\sqrt{1-\frac{1}{n^{2}}}$

Option 4)
$\sqrt{\frac{1}{1-n^{2}}}$

Answers (2)

As we learned

Acceleration of Block sliding down over rough inclined plane -

$ma=mg\ sin \theta - F$

$ma=mg\ sin \theta - \mu R$

$ma=mg\ sin \theta - \mu mg cos\theta$

$a=[sin \theta-\mu cos \theta]$

a = acceleration

$\mu=$ coefficient

- wherein

For $\mu=0$

$\therefore\ \;a=g\;sin\theta$

for smooth place

a = given

$s=\frac{1}{2}$ given $t^{2} \Rightarrow t_{1} = \sqrt{\frac{25}{g\sin \theta }}$

for rough surface

a = $g(\sin \theta -\cos \theta )$

$s = \frac{1}{2}g(\sin \theta -\mu \cos \theta )t_{2}^{2} or t_{2} = \sqrt{\frac{25}{g(\sin \theta -\mu \cos \theta )}}$

$\because t_{2} n t_{1}$

$\sqrt{\frac{25}{g(\sin \theta -\mu \cos \theta )}} = n.\sqrt{\frac{25}{g\sin \theta }}$

or $\sin \theta =n^{2}(\sin \theta -\mu \cos \theta )$

$\sin \theta =\cos \theta =\frac{1}{\sqrt{2}}\Rightarrow 1=n^{2}(1-4)$

$\mu - 1-\frac{1}{n^{2}}$

Option 1)

$\frac{1 }{2-n^2}$

Option 2)

$1-\frac{1}{n^{2}}$

Option 3)

$\sqrt{1-\frac{1}{n^{2}}}$

Option 4)

$\sqrt{\frac{1}{1-n^{2}}}$

Posted by

Vakul

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