I was hoping somebody could check over my work and see if I have the correct answer for this problem, as my text hasn't been the greatest in explaining this.(adsbygoogle = window.adsbygoogle || []).push({});

QUESTION

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

b) the distance travelled down the hill

MY ANSWER

Fa = Fgh - Ff

= Fg sin 5(degrees) - ukFn

= mg sin 5(degrees) - uk mg cos 5(degrees)

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

= -27.5N

F = ma

a = -27.5N / 25kg

a = -1.1m/s^2

delta t = (vf - vi) / a

= (0 - 3.5m/s) / -1.1m/s^2

= 3.2s

a) Therefore it takes the skier 3.2 seconds to come to a stop.

delta d = vi(t)+0.5(a)(t)^2

= (3.5 m/s)(3.2s) + 0.5(-1.1m/s^2)(3.2s)^2

= 11.2m - 5.632m

= 5.568m

= 5.6m

b) Therefore the skier travelled 5.6 meters down the hill.

Thanks so much for your time, I really appreciate it!

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# Homework Help: Kinetic Friction, did I do this properly?

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