Two ball of same mass are projected as shown. By compressing equally (say `x`) the springs of different force constants `K_(1)` and `K_(2)` by equal magnitude. The first ball is projected upwards along smooth wall and the other on the rough horizontal floor with coefficient of friction `mu`. If the first ball goes up by height `h`, then the distance covered by the second ball will be :
.

Options

A. `(2hK_(2))/(mu K_(1))`
B. `(hK_(1))/(2 mu K_(2))`
C. `(3h K_(2))/(2 mu K_(1))`
D. `(h K_(2))/(mu K_(1))`

Answer

D

Explanation

`(1)/(2)K_(1)x^(2) =mgh` (or) `x^(2)=(2mgh)/(K_(1))`
`(1)/(2) K_(2)x^(2) = mu mg x_(0)`(or) `x^(2) = (2mu mg x_(0))/(K_(2))`
equating `x_(0)=(h)/(mu).(K_(2))/(K_(1))`.