Two balls of same mass are projected as shown, by compressing the springs of different force constants `K_(1)` and `K_(2)` by equal distance. 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))/(muK_(1))`
B. `(hK_(1))/(2muK_(2))`
C. `(3hK_(2))/(2muK_(1))`
D. `(hK_(2))/(muK_(1))`

Answer

4

Explanation

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