## Essential University Physics: Volume 1 (4th Edition) Clone

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: $\mu_{k1}=0.62$