Given data:
* The mass of the skier is m = 55 kg.
* The angle of the slope is,
[tex]\theta=25^{\circ}[/tex]
* The initial velocity of the skier at the top is,
[tex]u=3.8\text{ m/s}[/tex]
* The coefficient of friction between the ski boot and slope is,
[tex]\mu=0.14[/tex]
* The distance traveled by the skier along the inclined plane is,
[tex]S=57\text{ m}[/tex]
Solution:
The diagrammatic representation of the given case is,
The angle of the inclined plane with the vertical axis is,
[tex]\begin{gathered} \alpha=180^{\circ}-90^{\circ}-\theta \\ \alpha=90^{\circ}-\theta \end{gathered}[/tex]
Thus, the component of the weight of skier along the inclined plane is,
[tex]\begin{gathered} F_1=mg\cos (90^{\circ}-\theta)_{} \\ F_1=mg\sin (\theta) \end{gathered}[/tex]
where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} F_1=55\times9.8\times\sin (25^{\circ}) \\ F_1=227.79\text{ N} \end{gathered}[/tex]
The component of weight perpendicular to the inclined plane is,
[tex]\begin{gathered} F_2=mg\sin (90^{\circ}-\theta) \\ F_2=mg\cos (\theta) \end{gathered}[/tex]
Substituting the known values,
[tex]\begin{gathered} F_2=55\times9.8\times\cos (25^{\circ}) \\ F_2=488.5\text{ N} \end{gathered}[/tex]
From the diagram, the normal force acting on the skier is,
[tex]\begin{gathered} F_N=F_2 \\ F_N=488.5\text{ N} \end{gathered}[/tex]
The frictional force acting on the skier in terms of the normal force is,
[tex]\begin{gathered} F_r=\mu F_N \\ F_r=0.14\times488.5 \\ F_r=68.39\text{ N} \end{gathered}[/tex]
Thus, the net force acting on the skier down the inclined plane is,
[tex]\begin{gathered} F=F_1-F_r_{} \\ F=227.79-68.39 \\ F=159.4\text{ N} \end{gathered}[/tex]
According to Newton's second law, the acceleration of the skier is,
[tex]\begin{gathered} F=ma \\ a=\frac{F}{m} \\ a=\frac{159.4}{55} \\ a=2.9ms^{-2} \end{gathered}[/tex]
By the kinematics equation, the speed of the skier after moving to distance S is,
[tex]\begin{gathered} v^2-u^2=2aS \\ v^2-3.8^2=2\times2.9\times57 \\ v^2-14.44=330.6 \\ v^2=330.6+14.44 \end{gathered}[/tex]
By simplifying,
[tex]\begin{gathered} v^2=345.04 \\ v=18.6\text{ m/s} \end{gathered}[/tex]
Thus, the speed of the skier after moving 57 m downhill is 18.6 meters per second.
