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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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Verified Answer
The correct answer is:
$2 \tan \phi$
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,
$$
\begin{aligned}
& v=0 \text { and } a=g(\sin \phi-\mu \cos \phi), \\
& \Rightarrow \quad 0=u^2+2 g(\sin \phi-\mu \cos \phi) \cdot \frac{l}{2} \\
& \Rightarrow-g l \sin \phi=g l(\sin \phi-\mu \cos \phi) \\
& \Rightarrow \quad \mu \cos \phi=2 \sin \phi \quad \Rightarrow \mu=2 \tan \phi
\end{aligned}
$$
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,
$$
\begin{aligned}
& v=0 \text { and } a=g(\sin \phi-\mu \cos \phi), \\
& \Rightarrow \quad 0=u^2+2 g(\sin \phi-\mu \cos \phi) \cdot \frac{l}{2} \\
& \Rightarrow-g l \sin \phi=g l(\sin \phi-\mu \cos \phi) \\
& \Rightarrow \quad \mu \cos \phi=2 \sin \phi \quad \Rightarrow \mu=2 \tan \phi
\end{aligned}
$$
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