The upper half of an inclined plane with an angle of inclination $\phi$, is smooth while the lower half is rough. A body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

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Aicharya · Accepted Answer

For upper half

From equation, $ v^2=u^2+2 a s $ we have, $u=0$ (from rest), $s=l / 2$ $ v^2=0+2(g \sin \phi) \cdot \frac{l}{2} $ For lower half, $ \begin{aligned} & v=0 \text { and } a=g(\sin \phi-\mu \cos \phi), \\ & \Rightarrow \quad 0=u^2+2 g(\sin \phi-\mu \cos \phi) \cdot \frac{l}{2} \\ & \Rightarrow-g l \sin \phi=g l(\sin \phi-\mu \cos \phi) \\ & \Rightarrow \quad \mu \cos \phi=2 \sin \phi \quad \Rightarrow \mu=2 \tan \phi \end{aligned} $

The upper half of an inclined plane with an angle of inclination $\phi$, is smooth while the lower half is…

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