Answer
Please see the work below.
Work Step by Step
We know that the acceleration of the slide is given as:
$a=gsin(\theta)-\mu_kgcos(\theta)$
and the time is given as:
$t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2d}{gsin(\theta)-\mu_kgcos(\theta)}}$
The ratio is:
$\frac{t_2}{t_1}=\frac{1}{3}=\frac{\sqrt{gsin(\theta)-\mu_{k1}gcos(\theta)}}{\sqrt{gsin(\theta)-\mu_{k2}gcos(\theta)}}$
We plug in the known values to obtain:
$\frac{t_2}{t_1}=\frac{1}{3}=\frac{\sqrt{(9.8)sin(35)-\mu_{k1}(9.8)cos(35)}}{\sqrt{(9.8)sin(35)-(0)(9.8)cos(35)}}$
$\implies \frac{1}{3}=\frac{\sqrt{5.621-(8.028)(\mu_{k1})}}{\sqrt{5.621}}$
This simplifies to:
$\mu_{k1}=0.62$
