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A small block starts slipping down from a point B on an inclined plane AB, which is making an angle $\theta$ with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by $\mu=k\tan\theta$. The value of k is .........

Numerical Value Type

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Ans. 3

Sol. Let $AC=l$ 

$\therefore BC=2l$ 

$\therefore AB=3l$

Apply work-Energy theorem

$W_f+W_{mg}=\Delta KE$

$mg(3l)\sin\theta-\mu mg\cos\theta(l)=0+0$

$\mu mg\cos\theta l=3mgl\sin\theta$

$\mu=3\tan\theta=k\tan\theta$

$\therefore k=3$

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You make it very easy

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