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?

 m

Jun 20th, 2015

Thank you for the opportunity to help you with your question!

mass=62 kg

friction coefficient =0.16

Q=mL

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

Q=2.5*3*10^5

Q=7.5*10^5 j



so the friction energy can find as fallows

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

but 

F=friction coefficient* R

R-the reactant force

so

F=0.16*62*9.81

F=97.32N

subustitute for (1)

W=97.32*S


but W=work=Q

so W=Q=7.5*10^5j

then

7.5*10^5=97.32*S

s=7.5*10^5/97.32

s=7706.5 m

so he has to go 7706.5 m

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 20th, 2015

please see the attach


ss 

Jun 21st, 2015

it is ok

Jun 21st, 2015

yes i re check that. it should be

Jun 21st, 2015

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Jun 20th, 2015
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