A body of mass m moving initially with velocity u on a rough surface and due to friction, it comes to rest after covering a distance s, then the retarding force is given by

$ma=\mu mg \mathrm{o}\mathrm{r}\mathrm{ }\mathrm{a}=\mu g$ .... (i)

Using, $v^2=u^2-2as,$ we have

$0=u^2-2\left(\mu g\right) s$ .... (ii)

$T=\frac{1}{2} m{u}^2\mathrm{o}\mathrm{r}\mathrm{ }{u}^2=\frac{2T}{m}$

We have from equation (ii),

$0=\frac{2T}{m}-\left(\mu g \right)s$

$\frac{2T}{m}=2\mu gy$ $\left[\because s=y\right]$ .... (iii)

Frictional force, $f=\mu mg$ .... (iv)

From equation (iii) and (iv), we get $f=\frac{T}{y}$