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  4. A block of mass m slides down along the smooth surface of the bowl (radius R ) from the rim to the bottom. The velocity of the block at the bottom will be

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Q. A block of mass $m$ slides down along the smooth surface of the bowl (radius $R$ ) from the rim to the bottom. The velocity of the block at the bottom will bePhysics Question Image

Work, Energy and Power

Solution:

Gain in $KE$. Loss in PE
$\frac{1}{2} m v^{2}=m g R \Rightarrow v=\sqrt{2 g R}$