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The upper half of an inclined plane with inclination $\phi$ is perfectly smooth while the lower half is rough .A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

Option 1)
$2\tan \phi$

Option 2)
$\tan \phi$

Option 3)
$2\sin \phi$

Option 4)
$2\cos \phi$

Answers (1)

best_answer

As we learnt in

Kinetic or Dynamic Friction -

$f_{K}\;\alpha\ R$

$f_{K}=\mu_{K} R$

$f_{K}=$ kinetic friction

$\mu_{K}=$ coefficient of kinetic friction

R = reaction

- wherein

$f_{K}<F_{l}$

$\therefore\ \mu_{K}<\mu_{s}$

$\mu_{K}=$ depends on the nature of surface in contact.

For upper half smooth incline, component of g down the incline = $gsin\phi$

$\therefore \; \; v^{2}=2(gsin\phi )\frac{l}{2}$

For lower half rough incline, frictional retardation = $\mu _{k}gcos\phi$

$\therefore$ Resultant acceleration = $gsin\phi -\mu _{k}gcos\phi$

$\therefore \; \; \; \; \; 0=v^{2}+2(gsin\phi -\mu _{k}gcos\phi )\frac{l}{2}$

or $0=2(gsin\phi )\frac{l}{2}+2g(sin\phi -\mu _{k}cos\phi )\frac{l}{2}$

$or\; \; \; 0=sin\phi +sin\phi-\mu _{k}cos\phi$

$or\; \; \; \mu _{k}cos\phi =2sin\phi$

$or\; \; \; \mu _{k}=2tan\phi$

Correct option is 1.

Option 1)

$2\tan \phi$

This is the correct option.

Option 2)

$\tan \phi$

This is an incorrect option.

Option 3)

$2\sin \phi$

This is an incorrect option.

Option 4)

$2\cos \phi$

This is an incorrect option.

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prateek

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