Answer :
We are given the following information
A skier is at rest on a hill sloped at 40°
The coefficient of kinetic friction between the snow and the skis is 0.12
How fast is the skier going after six seconds?
Let us first draw a free body diagram to better understand the problem
The sum of forces in the vertical direction is given by
[tex]\begin{gathered} \sum F_y=0 \\ F_N-F_g\cos \theta=0 \\ F_N=F_g\cos \theta \end{gathered}[/tex]The sum of forces in the horizontal direction is given by
[tex]\begin{gathered} \sum F_x=ma_x \\ F_g\sin \theta-F_k=ma_x \\ a_x=\frac{F_g\sin\theta-F_k}{m} \end{gathered}[/tex]The force due to friction is equal to
[tex]F_k=\mu_kF_N=\mu_kF_g\cos \theta[/tex]Where μk is the coefficient of kinetic friction between the snow and the ski.
[tex]a_x=\frac{F_g\sin\theta-\mu_kF_g\cos\theta}{m}[/tex]Substituting F = mg
[tex]a_x=\frac{mg\sin\theta-\mu_kmg\cos\theta}{m}=\frac{m(g\sin \theta-\mu_kg\cos \theta)}{m}[/tex]Mass cancels out
[tex]\begin{gathered} a_x=g\sin \theta-\mu_kg\cos \theta \\ a_x=9.81\cdot\sin (40\degree)-0.12\cdot9.81\cdot\cos (40\degree) \\ a_x=6.3-0.9 \\ a_x=5.4\; \; \frac{m}{s^2} \end{gathered}[/tex]Finally, since we have acceleration, we can find the velocity using the following equation of motion
[tex]v_f=v_i+a_x\cdot t[/tex]Where vf is the final velocity that we need to find out, vi is the initial velocity that is 0 since the skier was at rest initially and t is the time.
[tex]\begin{gathered} v_f=0+5.4\cdot6 \\ v_f=32.4\; \; \frac{m}{s} \end{gathered}[/tex]Therefore, the skier is going at 32.4 m/s
