In downhill speed skiing a skier is retarded by both the air drag force on the b

Question

In downhill speed skiing a skier is retarded by both the air drag force on the body and the kinetic frictional force on the skis. Suppose the slope angle is , the snow is dry snow with a coeffcient of kinetic friction k , the mass of the skier and equipment is m, the cross-sectional area of the (tucked) skier is A, the drag coeffcient is C, and the air density is . (a) What is the terminal speed? (b) If a skier can vary C by a slight amount dC by adjusting, say, the hand positions, what is the corresponding variation in the terminal speed? Give your answer in terms of the given values and g (note: dC counts as a variable, independent of C).

Explanation / Answer

Vt is the terminal velocity of the skier. Answers are: a) Vt = v[2mg(sinT - µcosT)/(CA?)] = 59 m/s b) Since Vt = k*C^(-½), dVt/dC = -½kC^(-3/2) A skier would have unlimited increasing velocity if there were no frictional forces to slow him down. The forces that limit velocity to the skier are K , which is the kinetic friction of his skis on the snow, and C, which is the air drag to his velocity. Those two frictional forces limit the skier to a terminal velocity. Alter either frictional force and you will increase or decrease terminal velocity. That's why they said if you tuck in your hands you decrease C which will in turn raise the Vt, the terminal velocity

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