Skiing and the Non-Conservative Force

With ski mountains around the country opening for the
season, I began to think about the physics of skiing. For a skier, we know that Energy at the top
and bottom of the mountain is equal to the skier’s kinetic energy plus the
potential energy:

E

_{tot}= KE + PE
At the top of the mountain, the skier has no velocity, and
only gravitational potential energy:

E

_{top}= (1/2)mv^{2}+ mgh = (1/2)m(0)^{2}+ mgh = mgh
At the bottom of the mountain, the skier has no kinetic
energy or gravitational potential energy (if the height at the bottom is
arbitrarily defined as 0. So, the work
done by the non-conservative forces between the top and the bottom of the
mountain is equal to the gravitational potential energy at the summit:

W

_{NC}= ∆KE + ∆PE = (0-0) + (0 – mgh) = -mgh
For my home mountain, Stratton Mountain in Stratton,
Vermont, h (the vertical drop) is 611 m.
Therefore, the work done by the non-conservative force on a skier of
mass 70 kg is equal to:

W

_{NC}= mgh = (70 kg)(9.8 m/s^{2})(611 m) = 419,146 J
But where does all this energy go? Most of it is lost to friction when the skier
turns his/her edges into the snow to decelerate (some of it is also lost to air
resistance). This seemed like a lot of
energy per skier, so I asked myself the question: if all of this energy were
transferred to the snow, how much snow would each skier melt per run? I assumed that the snow was already at its
melting point, and used the enthalpy of fusion for water (334 J/g) to determine
that each run by a 70 kg skier could melt 1.25 kg of snow.

419,146 J x (1 gram snow/334 J) = 1254.9 g = 1.25 kg snow

If Stratton Mountain were operating at full capacity (33,428
skiers/hour), there would be a total of 4.2 x 10

^{7}kg of snow melted per hour. Thus, most of the energy from the non-conservative work must go elsewhere, and not directly to melting snow.
## No comments:

## Post a Comment