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A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long, and the coefficient of friction between snow and skis is 0.075. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?

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Lanuel

The distance the skier glide along the horizontal portion of the snow before coming to rest is 289.31 meters.

Given the following data:

  • Initial velocity = 0 m/s (since the skier starts from rest).
  • Angle of inclination = 10.5°
  • Distance = 200 meter
  • Coefficient of friction = 0.075

To find how far (distance) the skier glide along the horizontal portion of the snow before coming to rest:

The height of the skier is given by the horizontal component of the hillside:

[tex]h_y = dsin(\theta)\\\\h_y = 200sin(10.5)\\\\h_y = 200\times0.1822[/tex]

Height, [tex]h_y[/tex] = 36.45 meters

For the horizontal component of the hillside:

[tex]h_x = dcos(\theta)\\\\h_x = 200cos(10.5)\\\\h_x = 200\times0.9833[/tex]

Height, [tex]h_x[/tex] = 196.65 meters

Next, we would determine the potential energy possessed by the skier:

[tex]P.E = mgh_y\\\\P.E = m \times 9.8 \times 36.45[/tex]

Potential energy = 357.18m Joules

Applying the law of conservation of energy:

The potential energy possessed by the skier is equal to the total sum of energy lost on the hill and flat surface due to friction:

[tex]P.E = E_h + E_f\\\\P.E = umgh_y + umgh_xx[/tex]

Substituting the given parameters into the formula, we have;

[tex]357.18m = 0.075\times m\times 9.8\times 196.65 + [0.075\times m\times9.8\times x]\\\\357.18m = 144.54m + 0.735mx\\\\0.735mx = 357.18m -144.54m\\\\0.735mx = 212.64m\\\\0.735x = 212.64\\\\x = \frac{212.64}{0.735}[/tex]

x = 289.31 meters

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