How Long Does It Take for a Skier to Stop on a Bunny Hill?

[quickslant]

A 25kg skier on a "bunny" hill that is inclined at 5 degrees has an intial velocity of 3.5 m/s. The coefficient of kinetic friction is 0.20

a) calculate the time taken for the skier to come to a stop
b) the distance traveled down the hill..

what i have so far is..

ive calculated is
Fg =mg = 25*(9.8) = 245N
Fn = mgcos 5 = 25*(9.8) * cos 5 = 244.07N
Friction = (mu) Fn = 0.20* 244.07 = 48.81N
F=ma thus 48.81 = 25a thus a= 1.95 m/s^2

a = Vf - Vi / delta T ... delta T = 1.79s

am i right so far?

 

[Saketh]

Check your net force equation. According to yours, friction is the only force acting on the skier. This would mean that he is being pushed by friction and friction only.

 

[quickslant]

ok so i have force normal, force of gravity and fictional force acting on the skier, so how do i add up those three?

 

[civil_dude]

You add up the ones parallel to the direction of acceleration. The normal force is not relevant, but the other component that makes up mg is relevant.

 

[quickslant]

so what i have now is Fg + Fk - mg sin 5 = net force
is what i have now correct?