Answer to Question #102185 in Mechanics | Relativity for dfdd

Question

A 62.7 kg skier rides a 916.6 m long lift to the top of a mountain. The lift makes an angle of 28.3o with the horizontal. How fast will she be traveling when she reaches the bottom of the mountain?

Answer 1

If we neglect friction, all we have to do is to use the conservation of energy: potential energy converts to kinetic energy. At the highest point the potential energy is

$$PE=mgH=mg\cdot L\text{ sin}\theta.$$

At the bottom this energy will turn into kinetic energy:

$$KE=\frac{1}{2}mv^2.$$

Since

$$PE=KE,\\ \space\\ mg\cdot L\text{ sin}\theta=\frac{1}{2}mv^2,\\ \space\\ v=\sqrt{2gL\text{ sin}\theta}=\\=\sqrt{2\cdot9.81\cdot916.6\cdot\text{sin}28.3^\circ}=92.3\text{ m/s}.$$

As we see, the final velocity does not depend on the mass the skier.