Yet another block sliding problem

Let's assume that the block's mass is equal to $m$

• Weight component tangent to the surface = $\frac{m}{\sqrt{2}}$

• Weight component normal to the surface = $\frac{m}{\sqrt{2}}$

In case of a smooth surface:

$\frac{mg}{\sqrt{2}} = ma$

$a = \frac{g}{\sqrt{2}}$

$\frac{d}{\sqrt{2}} = \frac{a t_1^2}{2}$

$\color{#3D99F6}{ t_1^2 = \frac{2\sqrt{2} d}{g} }$

In case of friction:

$\frac{mg}{\sqrt{2}} - \frac{mg \mu}{\sqrt{2}} = ma$

$a = \frac{g}{\sqrt{2}} (1- \mu)$

$\frac{d}{\sqrt{2}} = \frac{a t_2^2}{2}$

$\color{#3D99F6}{ t_2^2 = \frac{2\sqrt{2} d}{g(1- \mu)} }$

Dividing out:

$\frac{t_2^2}{t_1^2} = n^2$

$\frac{1}{1 - \mu} = n^2$

$\color{#D61F06}{\mu = 1 - \frac{1}{n^2}}$