http://skiingchampion.com/extreme-skiing/an-extreme-skier-starting-from-rest-coasts-down-a-mountain-that-makes-an-angle Best Skiing News and Resources Wed, 03 Jun 2009 14:50:20 +0000 hourly 1 http://wordpress.org/?v=3.3 http://skiingchampion.com/extreme-skiing/an-extreme-skier-starting-from-rest-coasts-down-a-mountain-that-makes-an-angle#comment-1410 Amy F Wed, 17 Dec 2008 21:39:59 +0000 http://skiingchampion.com/extreme-skiing/an-extreme-skier-starting-from-rest-coasts-down-a-mountain-that-makes-an-angle#comment-1410 First, find the acceleration of the skier. Draw a free body diagram. Net force = ma = mgsinΘ - umgcosΘ a = g(sinΘ - ugcosΘ) = 9.8(sin25 - 0.2cos25) = 2.37 m/s² Find speed at the end of 8.6 meters: v² = ax = 2.37 * 8.6 = 20.382 v = 4.51 m/s Now use conservation of energy: ½mvo² + mgh = ½mvf² ½vo² + gh = ½vf² ½*4.51² + 9.8*2.8 = ½*vf² vf = 8.67 m/s<br><b>References : </b><br> First, find the acceleration of the skier.
Draw a free body diagram.
Net force = ma = mgsinΘ – umgcosΘ
a = g(sinΘ – ugcosΘ) = 9.8(sin25 – 0.2cos25) = 2.37 m/s²
Find speed at the end of 8.6 meters:
v² = ax = 2.37 * 8.6 = 20.382
v = 4.51 m/s
Now use conservation of energy:
½mvo² + mgh = ½mvf²
½vo² + gh = ½vf²
½*4.51² + 9.8*2.8 = ½*vf²
vf = 8.67 m/s
References :

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