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(A) Let the total length of the inclined plane be $l$. The upper half has length $l/2$ and is smooth $(\mu = 0)$. The lower half has length $l/2$ and

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$2 	an \phi$

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(A) Let the total length of the inclined plane be $l$. The upper half has length $l/2$ and is smooth $(\mu = 0)$. The lower half has length $l/2$ and is rough with coefficient of friction $\mu$.For the upper half:The acceleration is $a_1 = g \sin \phi$. Starting from rest $(u = 0)$, the velocity $v$ at the midpoint is given by $v^2 = u^2 + 2 a_1 s$, where $s = l/2$.$v^2 = 0 + 2(g \sin \phi) \cdot (l/2) = gl \sin \phi$.For the lower half:The initial velocity is $v$, and the final velocity is $0$. The acceleration is $a_2 = g(\sin \phi - \mu \cos \phi)$. Using $v_f^2 = v_i^2 + 2 a_2 s$, where $s = l/2$:$0 = v^2 + 2g(\sin \phi - \mu \cos \phi) \cdot (l/2)$.Substituting $v^2 = gl \sin \phi$:$0 = gl \sin \phi + gl(\sin \phi - \mu \cos \phi)$.$0 = gl \sin \phi + gl \sin \phi - gl \mu \cos \phi$.$gl \mu \cos \phi = 2gl \sin \phi$.$\mu = 2 \frac{\sin \phi}{\cos \phi} = 2 	an \phi$.

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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