- #1
quickslant
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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?
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?