Question

Starting from rest, a skier slides 100 m down a 28 degree slope.How much longer does the run take if the coefficient of kinetic friction is 0.17 instead of 0?
Case 1: zero friction
F=mg sin(28)
a=F/m=g sin(28)
S=ut+(1/2)at²
100=0+g sin(28)t²/2
t=sqrt(2*100/(g sin(28))
=6.59 s.

Case 2: μ<sub>k</sub>=0.17
Frictional force
=μmg cos(28)
Net force (assuming μcos<sin)
=mg(sin(28)-μcos(28)
t=sqrt(2*100/(g (sin(28)-μcos(28))
=7.99 s

Take the difference.
shouldn't case 1 be cos(28) lol
I didn't understand where the net force equation in case 2 came from
nvm I got it

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