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:
Etot = KE + PE
At the top of the mountain, the skier has no velocity, and only gravitational potential energy:
Etop = (1/2)mv2 + 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:
WNC = ∆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:
WNC = mgh = (70 kg)(9.8 m/s2)(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 107 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.