A small block starts sliding down an inclined plane forming an angle `45^(@)` horizontal. The coefficient of friction `mu`, varies with distence s as `mu = cs^(2)` wherem, c is a constant of appropriate dimension, then distence covered by the block before it stops is

Options

A. `sqrt(3/c)`
B. `sqrt(3C)`
C. `sqrt(C)`
D. `sqrt((1)/(C))`

Answer

A

Explanation

Given that,

`mu=Cs^2" "{ :.s = "distance and C = constant")`
Net force o9n the block is

`Mgsin theta0 f= Ma`
`Mg sin theta- mu Mg cos theta = ma`
`g sin theta - mu g costheta= a`
`g[sin theta - Cs^2 costheta]=a`
`:." " a = (dv)/(dt) = v(dv)/(ds)`
`ads = vdv`
From Eqs. (i) and (ii), we get
`rArr" "g[ sin theta-Cs^2costheta]ds=vdv`
Integrating both sides, we get
`rArr " "(g sin theta)s - g (Cs^3)/3 costheta= v^2/2 + K ...(iii)`
For `theta= 45^@`, we have,
`(g/sqrt2\)s- ((Cs^2)/3)g/sqrt2=v^2/2+K`
Initially `t = 0, s = 0, u = 0`, substituting these, we get k = 0
So, `g /sqrt2(s-(Cs^2)/3= v^2/2 xx sqrt2/g)`
`rArr s = (Cs^3)/3 = v^2/2 xx sqrt2/g`
The body stops, when v = 0 `rArr s-(Cs^3)/3 = 0`
`rArr s = (Cs^3)/3 rArr s = sqrt(3/C)`