A skier slides down a hill, starting from rest at its top. The hill is shaped like an inclined plane that is 135 m long and inclined at an angle 17 degrees relative to the horizontal. If there is a coefficient of kinetic friction of 0.097 between the skier and the hill, how long does it take them to reach the bottom? (g = 9.8 m/s).

While the friction on the skier can be found from the normal force on the skier, the magnitude of this normal force is not directly given. To find the normal force on the skier, make use of the fact that there is no motion in the direction perpendicular to the inclined plane, such that forces in that direction should be balanced.

Let [tex]\theta = 17^{\circ}[/tex] denote the angle of elevation of the inclined plane. This angle is also the angle between the weight of the skier and the direction that is perpendicular to the inclined plane. Decompose the forces to find its component perpendicular to and in the direction of the inclined plane.

Perpendicular to the inclined plane:

Weight: [tex]m\, g\, \cos(17^{\circ})[/tex] since this angle is adjacent to the perpendicular component of this force.
Normal force: this force is entirely perpendicular to the inclined plane, so the magnitude of normal force would be the same as the magnitude of the component in this direction.
Friction: [tex]0[/tex], since friction is parallel to the inclined plane and is entirely contained in that plane.

In this direction of the inclined plane (parallel to the inclined plane):

Weight: [tex]m\, g\, \sin(17^{\circ})[/tex] since this angle is opposite the parallel component of this force.
Normal force: [tex]0[/tex], since this force is perpendicular to the inclined plane.
Friction: same as the magnitude of the entire force, since this force is parallel to this direction.

Using the fact that forces should be balanced in the direction perpendicular to the inclined plane, the magnitude of the normal force on the skier would be equal to that of the perpendicular component weight: [tex]m\, g\, \cos(17^{\circ})[/tex].

Let [tex]\mu_{k} = 0.097[/tex] denote the coefficient of static friction. While the skier is moving relative to the plane, the kinetic friction on the skier would be the product of the coefficient of kinetic friction and the normal force between the two two surfaces:

[tex](\mu_{k})\, (\text{normal force}) = (0.097)\, m\, g\, \cos(17^{\circ})[/tex].

Because forces in the perpendicular direction are balanced, the net force on the skier would be the combined effect of forces in the parallel direction- namely friction and the parallel component of weight. In this case, weight of the object (the component parallel to the incline) is in the direction of the motion while friction opposes the motion. Hence, the resultant force would be:

[tex]\begin{aligned} & (\text{net force on the skier}) \\ =\; & (\text{magnitude of weight, parallel component}) \\ & - (\text{magnitude of friction}) \\ =\; & m\, g\, \sin(17^{\circ}) - (0.097)\, m\, g\, \cos(17^{\circ}) \\ =\; & m\, g\, \left(\sin(17^{\circ}) - (0.097)\, \cos(17^{\circ})\right)\end{aligned}[/tex].

To find acceleration of the skier, divide net force by mass:

[tex]\begin{aligned} & (\text{acceleration}) \\ =\; & \frac{(\text{net force})}{(\text{mass})} \\ =\; & \frac{(m\, g)\, \left(\sin(17^{\circ}) - (0.097)\, \cos(17^{\circ})\right)}{m} \\ =\; & (g)\, \left(\sin(17^{\circ}) - (0.097)\, \cos(17^{\circ})\right) \\ =\; & (9.8\; {\rm m\cdot s^{-2}})\, \left(\sin(17^{\circ}) - (0.097)\, \cos(17^{\circ})\right) \\ \approx\; & 1.95618\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].

Make use of the SUVAT equations to find the time required. In this motion:

Acceleration is constantly [tex]a \approx 1.95618\; {\rm m\cdot s^{-2}}[/tex]
Initial velocity is [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex].
Displacement is [tex]x = 135\; {\rm m}[/tex].

Start by finding the velocity [tex]v[/tex] right after the acceleration using the equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex]:

[tex]\begin{aligned}v &= \sqrt{u^{2} + 2\, a\, x}\end{aligned}[/tex].

To find the duration [tex]t[/tex] of the motion, divide the change in velocity [tex](v - u)[/tex] by acceleration:

[tex]\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{\left(\sqrt{u^{2} + 2\, a\, x\right) - u}}{a} \\ &\approx \frac{\left(\sqrt{0^{2} + 2\, (1.95618)\, (135)}\right) - 0}{1.95618}\; {\rm s} \\ &\approx 11.7\; {\rm s}\end{aligned}[/tex].

In other words, the motion would require approximately [tex]11.7\; {\rm s}[/tex].