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 cross-sectional 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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