A small bead of mass m is placed at the bottom of watch glass of radius R. It is displaced by h (h << R) along the glass surface and released. Calculate the total distance described by it before it comes to rest at the bottom. $\mathrm{\mu }$ is the coefficient of friction between bead and the watch glass.

 

Since gravity is a conservation force work done by it depends only on the initial and final position and not on the length of the path. But work done by friction depends on the length of path as it is non-conservative.

${\mathrm{W}}_{\mathrm{mg}} = \mathrm{mgh}$
Here $\mathrm{N} = \mathrm{mg} \cos \mathrm{\theta } = \mathrm{mg}(\therefore \mathrm{\theta } \mathrm{is} \mathrm{small})$
${\mathrm{F}}_{\mathrm{friction}}= \mathrm{\mu N} = \mathrm{\mu mg}$
Work done by friction = - $\mathrm{\mu mgs}$, where s is the length
of path
By work--energy theorem,
mgh - $\mathrm{\mu }$mgs = 0 [zero because final kinetic energy is zero and initial kinetic energy is also zero]

$\therefore \mathrm{s} = \frac{\mathrm{h}}{\mathrm{\mu }}$