A skier of mass 70kg skies down a hill angled at 35 degrees. The coefficient of
ID: 1305721 • Letter: A
Question
A skier of mass 70kg skies down a hill angled at 35 degrees. The coefficient of friction between her skies and the snow is muk=.1 and she travels a total distance of 12m along the hill during her trip. She begins at rest. Assume the skier does not use her poles or have any other forces on her. a) how much work does the force of gravity do on the skier? b) How much work does the normal force do on the skier? c) How much work does the friction force do against the skier? d) The total work done on the skier is the sum of those three answers since those are the only forces on the skier. Use the fact that sigma W=delta K to determine the skier's final speed.
Explanation / Answer
a) Wg = Fg*d*cos(90-35)
= m*g*d*sin(35)
= 70*9.8*12*sin(35)
= 4721.68 J
b) WN = 0
c) Wf = -mue*m*g*cos(35)*d
= -0.1*70*9.8*cos(35)*12
= -674.33 J
d) Wnet = Wg + WN + Wf
= 4721.68 + 0 - 674.33
= 4047.35 J
we know, Wnet = 0.5*m*v^2
v = sqrt(2*Wnet/m)
= sqrt(2*4047.35/70)
= 10.735 m/s
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