A 90 kg skier competing in the Nordic Combined event travels down a parabolic sk

Question

A 90 kg skier competing in the Nordic Combined event travels down a parabolic ski ramp described by the equation y = 0.01 x^2 - 0.1x + 0.25, where the origin is the point on the ground below the end of the ski ramp as shown in the figure. He starts from rest at the point x = 30m and simply slides (he does not push himself with his ski poles). If the friction with the snow is negligible, what is the magnitude of his velocity when he leaves the ramp? If the force of friction is not negligible, the snow exerts a Force mu F_N t, where t is a unit vector tangent to the skier's trajectory (tangent to the ski slope). For the final portion of the ramp (where the skier is going back up for liftoff after hitting a height of zero), is the snow doing positive or negative work on the skier? Explain.

Explanation / Answer

Y = 0.01x^2 -0.1x + 0.25

for X=30

Y = 0.01*(30)^2 - 0.1*30 + 0.25 = 6.25 m

energy conservation

loss in PE = gain in KE

mgh = 1/2mV^2

=> V = sqrt(2gh)

=> V = sqrt(2*9.8*6.25) = 11.06 m/s

b) friction always do negative work ( as friction act opposite to motion)

=> snow will do -tive work on skier

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