# A 62-kg cross-country skier glides over snow as in the figure below. The coeffic

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62-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°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 2.5 kg of snow?

Jun 21st, 2015

mass=62 kg

friction coefficient =0.16

Q=mL

L-latent heat for ice=3*10^5 j/kg

Q=2.5 x 3 x 10^5

Q=7.5 x 10^5 j

so the friction energy can find as fallows

W=Fs-------------(1)

but

F=friction coefficient x R  where  R is the reactant force

so

F=0.16 x 62 x 9.81

F=97.32N

subustitute for (1)

W=97.32 x S

but W=work=Q

so W=Q=7.5 x 10^5j

then

7.5 x 10^5=97.32 x S

s=7.5 x 10^5/97.32 which is

s=7706.5 m

so he has to ski 7706.5 m

Thank you.
Jun 21st, 2015

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Jun 21st, 2015
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Jun 21st, 2015
Nov 19th, 2017
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