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(adsbygoogle = window.adsbygoogle || []).push({});

a) calculate the time taken for the skier to come to a stop

b) the distance travelled 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?

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# Another friction question

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