A 64-kg cross-country skier glides over snow as in the figure below. The coeffic

Question

A 64-kg cross-country skier glides over snow as in the figure below. The coefficient of friction between skis and snow is 0.16. Assume all the snow beneath her skis is at 0 A 64-kg cross-country skier glides over snow as in the figure below. The coefficient of friction between skis and snow is 0.16. Assume all the snow beneath her skis is at 0 A 64-kg cross-country skier glides over snow as in the figure below. The coefficient of friction between skis and snow is 0.16. Assume all the snow beneath her skis is at 0 Degree C and that all the internal energy generated by friction is added to snow, which sticks to her skis until it melts. How far would she have to ski to melt 2.0 kg of snow? M Read It Need Help?

Explanation / Answer

Heat generated by friction = Work by friction = Heat for melting
Force of friction * Distance = Mass * Latent Heat of Fusion
(Coefficient)(Mass of skier)(9.8m/s^2)(Distance) = 2kg*(334J/kg)
(0.16)(64kg)(9.8m/s^2)(d) = 668J
d = 6.66 m

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