Question

A skier traveling 11.0 m/s reaches the foot of a steady upward 17 â incline and glides 15 m up along this slope before coming to rest. part a what was the average coefficient of friction?

Answer 1

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 $mgSin \alpha$ 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 $F_{net} = ma$

The $F_{net}$ here is given by $mgSin \alpha + F_{k}$, where $F_{k}$ is the Kinetic Friction.

We also know that the magnitude of Friction Force can be calculated using the equation $F_{k} =$ μ.$F_{N}$, where $F_{N}$ is the Normal Force acting perpendicular to the incline as shown in the figure.

We see that $F_{N} = mgCos \alpha$ since they both balance each other out.

Hence, putting all these together, we have $mgSin \alpha +$μ.$mgCos \alpha = ma$

Simplifying this, we get $gSin \alpha +$ μ$gCos \alpha = a$

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 $V_{i} = 11.0 m/s$

Final Velocity $V_{f} = 0$

Displacement along the incline $D = 15m$

Acceleration a = ?

Using the equation $V_{f} ^{2} = V_{i} ^{2} + 2aD$, and

Plugging in known numerical values, we get $0 = (11)^{2} + 2a(15)$

Solving for a gives us, $a = -4.03 m/s^{2}$

Since the negative sign indicates that this is deceleration, we can ignore the sign and consider the magnitude alone.

Thus, plugging in $a = 4.03 m/s^{2}$ in the force equation we wrote above, we have

$(9.8)Sin (17) +$ μ.$(9.8)Cos (17) = 4.03$

Solving this for μ, we get its value as μ = 0.124

Thus, the average coefficient of friction on the incline is 0.12