Question

Calculate the acceleration of a skier heading down a 10.0º slope, assuming the coefficient of friction for waxed wood on wet snow. (b) find the angle of the slope down which this skier could coast at a constant velocity. you can neglect air resistance in both parts, and you will find the result of exercise 5.9 to be useful. explicitly show how you follow the steps in the problem-solving strategies.

Answer 1

(a) 0.74 m/s^2

Explanation:

There are two forces acting on the skier: the component of the weight parallel to the slope, which acts downward, and the frictional force, which acts upward along the incline.

The component of the weight parallel to the inclined plane is:

$W= m g sin \theta$

where m is the mass of the skier, $g=9.81 m/s^2$ and $\theta=10^{\circ}$.

The frictional force is instead

$F_f = -\mu m g cos \theta$

$\mu=0.1$ is the coefficient of friction for waxed wood on wet snow.

If we apply Newton's second law, we can write that the net force must be equal to the product of mass per acceleration:

$mgsin \theta -\mu mg cos \theta =ma$

And symplifying m, we can find the acceleration:

$a=g sin \theta-\mu g cos \theta=$

$=(9.81 m/s^2)(sin 10^{\circ})-(0.1)(9.81 m/s^2)(cos 10^{\circ})=0.74 m/s^2$

(b) $5.7^{\circ}$

Explanation:

This time, the skier is moving at constant velocity. Therefore, the acceleration is zero (a=0) and Newton's second law becomes:

$mg sin \theta - \mu m g cos \theta=0$

By simplifying, we get

$tan \theta = \mu$

From which we can find the angle at which the skier could coast at a constant velocity:

$\theta= tan^{-1} (\mu) = tan^{-1} (0.1)=5.7^{\circ}$