As depicted below, a snow skier of mass 70.0 kg pushes off at the top of a hill
ID: 1996977 • Letter: A
Question
As depicted below, a snow skier of mass 70.0 kg pushes off at the top of a hill (point A) with an initial velocity of 2.00 m/s. Although she has waxed her skies, there is still some kinetic friction between her skies and the snow covered hill. When she is at point B her velocity is measured by "radar gun" and found to be 18.0 m/s. How much work was done by the force of kinetic friction between points A and B? Without having stopped at point B, she continues on her ski run. And after reaching the bottom of the hill she continues to glide up the hill before momentarily coming to rest at some maximum final elevation at point C. Somehow, some "smarty pants" CALC III student who is very good at evaluating Vector Line Integrals calculates that the total work done by friction between points B and C was -20,000 J (negative 20,000 J). What was her maximum final elevation at point C?Explanation / Answer
(a) Applying work-energy theorem,
Work done by gravity + work done by fricrtion = change in KE
(70 x 9.8 x (90 -50)) + (Wf) = 70 (18^2 - 2^2) / 2
27440 + Wf = 11200
Wf = - 16240 J .......Ans
(B) Work done by gravity + Work done by friction = change In KE
(70 x 9.8 x h ) + (- 20,000) = 0 - 70(2^2)/2
686h - 20000 = - 140
h = 28.95 m
y = 90 - h = 118.95 m
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