The attached free-body diagram shows the forces that are responsible for the skier coming to rest eventually on the incline.
We see that the component of his Weight along the incline [tex]mgSin \alpha[/tex] and the Friction both act in tandem to stop him.
In order to calculate the Friction, we can make use of Newton's 2nd law, which states that [tex]F_{net} = ma[/tex]
The [tex]F_{net}[/tex] here is given by [tex]mgSin \alpha + F_{k}[/tex], where [tex]F_{k}[/tex] is the Kinetic Friction.
We also know that the magnitude of Friction Force can be calculated using the equation [tex]F_{k} =[/tex] μ.[tex]F_{N}[/tex], where [tex]F_{N}[/tex] is the Normal Force acting perpendicular to the incline as shown in the figure.
We see that [tex]F_{N} = mgCos \alpha[/tex] since they both balance each other out.
Hence, putting all these together, we have [tex]mgSin \alpha +[/tex]μ.[tex]mgCos \alpha = ma[/tex]
Simplifying this, we get [tex]gSin \alpha +[/tex] μ[tex]gCos \alpha = a[/tex]
We clearly see that we need to calculate the acceleration before we can obtain the value of the Coefficient of Friction μ
And for that, we make use of the following data obtained from the question:
Initial Velocity [tex]V_{i} = 11.0 m/s[/tex]
Final Velocity [tex]V_{f} = 0[/tex]
Displacement along the incline [tex]D = 15m[/tex]
Acceleration a = ?
Using the equation [tex]V_{f} ^{2} = V_{i} ^{2} + 2aD[/tex], and
Plugging in known numerical values, we get [tex]0 = (11)^{2} + 2a(15)[/tex]
Solving for a gives us, [tex]a = -4.03 m/s^{2}[/tex]
Since the negative sign indicates that this is deceleration, we can ignore the sign and consider the magnitude alone.
Thus, plugging in [tex]a = 4.03 m/s^{2}[/tex] in the force equation we wrote above, we have
[tex](9.8)Sin (17) +[/tex] μ.[tex](9.8)Cos (17) = 4.03[/tex]
Solving this for μ, we get its value as μ = 0.124
Thus, the average coefficient of friction on the incline is 0.12
