In downhill speed skiing a skier is 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?

(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)

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  1. 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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  2. def class 张徐帅:
    def __init__(self, x, y):
    return self.x = x
    return self.y = y
    def ity(self):
    return self.x

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