##### A 62-kg cross-country skier glides over snow as in the figure below. The coefficient of friction bet

label Physics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

A 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? Incorrect: Your answer is incorrect. Determine the force of friction and require the force of friction to do the work that melts the ice. m

Oct 19th, 2017

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

Jun 21st, 2015

wrong

Jun 21st, 2015

Jun 21st, 2015

oh it can be any mistake in calculation...... solution    it is too lengthy ...... you can refund you 1\$

but not give negative review.... because your question's price is not 1\$ it is too lengthy ..

Thanks

best of luck

Jun 21st, 2015

...
Oct 19th, 2017
...
Oct 19th, 2017
Oct 20th, 2017
check_circle