Question

Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vertical cross-sectional area of 0.040 m^2 and has a drag coefficient C of 0.80. Take the air density to be 1.21 kg/m^3, and the coefficient of kinetic friction to be 0.80. (a) In kilometers per hour, what wind speed V along the ground is needed to maintain the stone’s motion once it has started moving?

Answer 1

Answer:

324.14 km/h

Explanation:

Data:

mass, m = 20 kg

vertical cross-sectional area, A = 0.040 m^2

drag coefficient, C = 0.80

air density, ρ = 1.21 kg/m^3

coefficient of kinetic friction, μ = 0.80

Eq. 6-14:

Drag-Force = (C*ρ*A*v^2)*(1/2)    (where v is wind speed)

But Drag-Force is also = m*g*μ  (where g is standard gravitational acceleration = 9.81 m/(s^2)). Therefore:

m*g*μ = (C*ρ*A*v^2)*(1/2)

Solving for v

v = √[m*g*μ*2/(C*ρ*A)]

v = √[20*9.81*0.8*2/(0.8*1.21*0.04)]

v = 90.04 m/s

To convert to km/h, multiply by 3.6

v = 90.04*3.6 = 324.14 km/h

Answer 2

Answer:

362.41 km/h

Explanation:

F = Force

m = Mass = 84 kg

g = Acceleration due to gravity = 9.81 m/s²

C = Drag coefficient = 0.8

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.04 m²

v = Terminal velocity

F = ma

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow mg=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{mg}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{20\times 9.81}{1.21\times 0.8\times 0.04}}\\\Rightarrow v=100.66924\ m/s$

Converting to km/h

$100.66924\times 3.6=362.41\ km/h$

The terminal velocity of the stone is 362.41 km/h