The upper half of an inclined plane with inclination φ is...

Acceleration of block while sliding down upper half $=\mathrm{g} \sin \phi$ retardation of block while sliding down lower half $=-(\mathrm{g} \sin \phi-\mu \mathrm{g} \cos \phi)$ For the block to come to rest at the bottom, acceleration in I half = retardation in II half. $\mathrm{g} \sin \phi=-(\mathrm{g} \sin \phi-\mu \mathrm{g} \cos \phi)\Rightarrow \mu=2 \tan \phi$ Alternative method: According to work-energy theorem, $\mathrm{W}=\Delta \mathrm{K}=0$ (Since initial and final speeds are zero) $\therefore$ Work done by friction $+$ Work done by gravity $=0$ i.e., $-(\mu \mathrm{mg} \cos \phi) \frac{\ell}{2}+\mathrm{mg} \ell \sin \phi=0$ or $\frac{\mu}{2} \cos \phi=\sin \phi$ or $\mu=2 \tan \phi$