17 Dec
An extreme skier, starting from rest, coasts down a mountain that makes an angle 25.0° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 8.6 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 2.80 m below the edge. How fast is she going just before she lands?
this answer is not correct… please help
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
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One Response for "An extreme skier, starting from rest, coasts down a mountain that makes an angle?"
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
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