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

label Physics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

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

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 20th, 2015

Jun 21st, 2015

it is ok

Jun 21st, 2015

yes i re check that. it should be

Jun 21st, 2015

...
Jun 20th, 2015
...
Jun 20th, 2015
Oct 21st, 2017
check_circle