A block of mass `m` has initial velocity `u` having direction towards `+x` axis. The block stops after covering a distnce S causing similar extension in the spring of constant K holding it. If `mu` is the kinetic friction between the block and the surface on which it was moving, the distance S is given

Options

A. (a) `1/Kmu^2m^2g^2`
B. (b) `1/K(mKu^2-mu^2m^2g^2)^(1//2)`
C. (c) `1/K(mu^2m^2g^2+mKmu^2+mumg)^(1//2)`
D. (d) `(-mumg+sqrt(mu^2m^2g^2+m u^2k))/(k)`

Answer

D

Explanation

By the work-energy theorem, `1/2m u^2=mumgS+1/2kS^2`
i.e., `S^2+(2mumgS)/(k)-(mu^2)/(k)=0`
`impliesS=((-2mumg)/(k)+sqrt((4mu^2m^2g^2))/(k^2)(4m u^2)/(k))/(2)`
`=(-mumg+sqrt(mu^2m^2g^2+m u^2k))/(k)`