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(A) Case $I$: As the block slides down with constant velocity,the acceleration is zero. In this case,the frictional force $f$ balances the component o

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$\frac{u^2}{4g \sin 	heta}$

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(A) Case $I$: As the block slides down with constant velocity,the acceleration is zero. In this case,the frictional force $f$ balances the component of gravity along the plane:$f = mg \sin 	heta$ and $f = \mu R = \mu mg \cos 	heta$Therefore,$\mu mg \cos 	heta = mg \sin 	heta$,which gives $\mu = 	an 	heta$ ... $(i)$Case $II$: The block is projected upward with initial velocity $u$. It experiences a downward acceleration $a$ due to both gravity and friction acting down the plane:$mg \sin 	heta + f = ma$$mg \sin 	heta + \mu mg \cos 	heta = ma$Substituting $\mu = 	an 	heta$ from $(i)$:$mg \sin 	heta + (	an 	heta) mg \cos 	heta = ma$$mg \sin 	heta + mg \sin 	heta = ma$$2mg \sin 	heta = ma \implies a = 2g \sin 	heta$ ... $(ii)$Let $x$ be the distance moved up the plane before the block comes to rest. Using the equation of motion $v^2 - u^2 = 2as$:$0^2 - u^2 = 2(-a)x$$-u^2 = -2(2g \sin 	heta)x$$x = \frac{u^2}{4g \sin 	heta}$

$A$ block slides down an inclined plane of slope angle $\theta$ with a constant velocity. It is then projected up the plane with an initial velocity $u$. The distance up to which it will rise before coming to rest is

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