 Friction on inclined plane of angle more than angle of repose
Question

# The upper half of an inclined plane of inclination $\theta$ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by

Moderate
Solution

## Let m be mass of the block and L be length of the inclined plane.            According to work-energy theorem         $W=\Delta K=0$         ($\because$Initial and final speeds are zero)       Work done by friction+ Work done by gravity= 0     Alternate solution           For upper half smooth plane Acceleration of the block, $a=g\mathrm{sin}\theta$Here, u = 0 ($\because$block starts from rest)$a=\mathrm{sin}\theta ,s=\frac{L}{2}$ Using, ${v}^{2}-{u}^{2}=2as,$we have     For lower half rough planeAcceleration of the block,${a}^{\text{'}}=g\mathrm{sin}\theta -\mu g\mathrm{cos}\theta$where $\mu$is the coefficient of friction between the block and lower half of the planeHere, $u=v=\sqrt{gL\mathrm{sin}\theta },$

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