A small head of `m` is pleced in the bottom of which glass of value radius `R` It is displaced by `(h ltlt R)` along the glass surface and released calculate the total distance described by it before it comes to rst of the botom `mu` is the coefficient of friction between bead and the watch glass

Options

A. `h//mu`
B. `h//2 mu`
C. `hsqrt(mu)`
D. `(h)/(mu R)`

Answer

A

Explanation

Since gravity is a cosrevation force work done by it depends only on the iniitial and final position and not on the length of the path .But work done by friction depends in the length of path as it is non consrvative

`W_(mg) = mgh`
Here `N = mg cos theta = mg (because theta is small)`
`F_("friction") = mu N = mu mg`
Work done by the friction `= - mu` mgs, where `s` is the length of path
`mgh - mu mgs = 0` (zero because final kinetic energy is zero and initial energy is also zero)
`:. s = (h)/(mu)`