Posted by johnny on Saturday, December 13, 2008 at 1:15pm.
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 è = 37.5°, the snow is dry snow with a coefficient of kinetic friction ìk = 0.0400, the mass of the skier and equipment is m = 89.5 kg, the crosssectional area of the (tucked) skier is A = 1.30 m2, the drag coefficient is C = 0.150, and the air density is 1.20 kg/m3.
(a) What is the terminal speed?
m/s
(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? (dvt/dC)
m/s

physics  drwls, Saturday, December 13, 2008 at 1:59pm
Compute the drag force as
(1/2)*C*(air density)*V^2*(Area)
Compute the drag force on the skis with the standard friction coeffient equation.
m g sin A  m g cos A Uk  (1/2) A C (density) V^2 = 0 = Fnet
(a) Write Newton's second law with zero acceleration. The only unknown will be Vt, the terminal speed. Solve for it
(b) Differentiate the law of motion to get dV(C)/dC. You will save a step if you do it impliclitly with respect to C
dFnet/dC = (1/2)A*density*V^2 + (1/2)*A*2V*C*dV/dC*(density)= 0
V^2 + 2 C V*dV/dC = 0
dV/dC = V/(2C)
2*dV/V = dC/C
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