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 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
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To find the terminal speed, we need to equate the forces due to air drag and kinetic friction.
The air drag force is given by the equation:
Drag force = (1/2) * air density * cross-sectional area * drag coefficient * velocity^2
where velocity is the speed of the skier.
The kinetic friction force is given by the equation:
Frictional force = coefficient of kinetic friction * mass * acceleration due to gravity
At terminal speed, the acceleration is zero. Therefore, the kinetic friction force is equal to the air drag force:
(coefficient of kinetic friction) * (mass) * (acceleration due to gravity) = (1/2) * (air density) * (cross-sectional area) * (drag coefficient) * (velocity^2)
Let's solve for the terminal speed.
(a) Terminal speed:
(coefficient of kinetic friction) * (mass) * (acceleration due to gravity) = (1/2) * (air density) * (cross-sectional area) * (drag coefficient) * (velocity^2)
Substituting the given values:
(0.0400) * (89.5 kg) * (9.8 m/s^2) = (1/2) * (1.20 kg/m^3) * (1.30 m^2) * (0.150) * (velocity^2)
Simplifying:
velocity^2 = (0.0400 * 89.5 * 9.8) / (0.5 * 1.20 * 1.30 * 0.150)
velocity^2 = 91.0618
velocity = √(91.0618) = 9.55 m/s (rounded to two decimal places)
So, the terminal speed is 9.55 m/s.
(b) To find the corresponding variation in the terminal speed, we can differentiate the equation with respect to the drag coefficient C:
(dFrictional force/dC) = (1/2) * (air density) * (cross-sectional area) * (dC/dt) * (velocity^2)
(dFrictional force/dC) = (1/2) * (1.20 kg/m^3) * (1.30 m^2) * (1) * (velocity^2)
Substituting the values calculated in part (a):
(dFrictional force/dC) = (1/2) * (1.20 kg/m^3) * (1.30 m^2) * (1) * (9.55^2)
(dFrictional force/dC) ≈ 70.04
So, the corresponding variation in terminal speed is 70.04 m/s for a slight variation in the drag coefficient.
To solve this problem, we need to find the values for the drag force and the kinetic frictional force, and then use them to calculate the terminal speed.
Step 1: Finding the drag force
We can calculate the drag force using the formula:
F_drag = (1/2) * ρ * A * C * v^2
where:
ρ is the air density (1.20 kg/m^3)
A is the cross-sectional area of the skier (1.30 m^2)
C is the drag coefficient (0.150)
v is the speed of the skier (variable)
Step 2: Finding the kinetic frictional force
We can calculate the kinetic frictional force using the formula:
F_friction = μk * m * g
where:
μk is the coefficient of kinetic friction (0.0400)
m is the mass of the skier and equipment (89.5 kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Step 3: Equate the drag force and the kinetic frictional force
At terminal speed, the drag force and kinetic frictional force are equal:
F_drag = F_friction
Step 4: Solve for the terminal speed
Setting F_drag = F_friction, we have:
(1/2) * ρ * A * C * v^2 = μk * m * g
Simplifying and solving for v, we get:
v^2 = (2 * μk * m * g) / (ρ * A * C)
Taking the square root of both sides:
v = √[(2 * μk * m * g) / (ρ * A * C)]
Now we can plug in the given values to calculate the terminal speed.
(a) Terminal Speed:
Plugging in the given values:
v = √[(2 * 0.0400 * 89.5 * 9.8) / (1.20 * 1.30 * 0.150)]
Calculating this value will give us the terminal speed in m/s.
(b) To find the corresponding variation in terminal speed with a slight change in the drag coefficient (dC), we can differentiate the terminal speed equation with respect to C:
(dv/dC) = -√[(2 * μk * m * g) / (ρ * A * C^3)]
This will give us the change in terminal speed for a small change in the drag coefficient.
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