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I'm not sure if I did this right but here is the word problem and the work i've done. If I made any errors, i'd love to know about them!

A sight seen on many bunny hills across Ontario is young skiers pushing on ski poles and gliding down a slope until they come to rest. Observing from a distance, you note a young person (approximately 25 kg) pushing off with the ski poles to give herself an initial velocity of 3.5 m/s. If the inclination of the hill is 5.0 degrees and the coefficient of kinetic friction for the skis on dry snow is 0.20, calculate

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

For a:

Fa = Fgh + Ff

= Fg sin 5 (degrees) + ukFn

Fn = Fg cos 5 (degrees)

= mg cos 5 (degrees)

Fa = Fg sin 5(degrees) + ukmg cos 5 (degrees)

= mg sin 5(degrees) + ukmg cos 5(degrees)

= (25kg) (9.8m/s^2)(0.08716) + (0.20)(25kg)(9.8m/s^2)(0.9962)

= 70N

F = ma

a = 70N / 25kg

= 2.8 m/s^2

delta t = (Vf - Vi) / a

= (0 - 3.5m/s) / 2.8 m/s^2

= 1.25s

The time for the skier to come to a stop is 1.25 seconds.

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# Homework Help: Kinetic friction - finding time

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