Fg = mg*sinθ - μk*Fn
= mg*sinθ - μk*mg*cosθ
= mg*(sinθ - μk*cosθ)
= 504.1426 (N)
a)
Vt = sqrt((2*Fg) / (C*P*A))
= 65.6 (m/s)
b)
dv = (-1/2)*sqrt((2*Fg) / (P*A))*C^(-3/2) dc
= -218.8 (m/s)
In downhill speed skiing a skier is ed by both the air drag force on the body and the kinetic frictional force on the skis. Suppose the slope angle is θ = 39.5°, the snow is dry snow with a coefficient of kinetic friction μk = 0.0400, the mass of the skier and equipment is m = 85.0 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
2 answers
Fbd is not there so may be the method u tried to solve it , could be wrong
2 answers