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
(a) Terminal speed is the maximum speed that a skier can reach while downhill skiing, where the drag force and the frictional force balance out. To find the terminal speed, we need to equate the two forces.
First, let's find the air drag force on the body of the skier. The formula for air drag force is:
F_drag = (1/2) * C * ρ * A * v^2
where:
C = drag coefficient
ρ = air density
A = cross-sectional area of the skier
v = velocity of the skier
Substituting the given values:
C = 0.150
ρ = 1.20 kg/m^3
A = 1.30 m^2
The air drag force is given by:
F_drag = (1/2) * 0.150 * 1.20 * 1.30 * v^2
Next, let's find the kinetic frictional force on the skis. The formula for kinetic frictional force is:
F_friction = μk * m * g
where:
μk = coefficient of kinetic friction
m = mass of the skier and equipment
g = acceleration due to gravity
Substituting the given values:
μk = 0.0400
m = 85.0 kg
g = 9.8 m/s^2
The kinetic frictional force is given by:
F_friction = 0.0400 * 85.0 * 9.8
Since the two forces balance out at terminal speed, we have:
F_drag = F_friction
Simplifying the equation:
(1/2) * 0.150 * 1.20 * 1.30 * v^2 = 0.0400 * 85.0 * 9.8
Solving for v^2:
v^2 = (0.0400 * 85.0 * 9.8) / ((1/2) * 0.150 * 1.20 * 1.30)
Now we can find the terminal speed:
v_terminal = sqrt[(0.0400 * 85.0 * 9.8) / ((1/2) * 0.150 * 1.20 * 1.30)]
Calculating the value will give you the terminal speed in m/s.
(b) To find the corresponding variation in the terminal speed with a slight variation in the drag coefficient (dC), we can differentiate the terminal speed equation with respect to C.
Differentiating the equation v_terminal = sqrt[(0.0400 * 85.0 * 9.8) / ((1/2) * 0.150 * 1.20 * 1.30)] with respect to C will give us:
(dvt/dC) = [sqrt[(0.0400 * 85.0 * 9.8) / ((1/2) * 0.150 * 1.20 * 1.30)]] * (1/2) * (0.0400 * 85.0 * 9.8) * -1 / ((1/2) * (0.150 * 1.20 * 1.30)^2)
Simplifying the equation will give you the corresponding variation in the terminal speed (dvt/dC) in m/s.
To find the terminal speed, we need to find the equilibrium between the gravitational force pulling the skier downhill and the drag force and frictional force opposing the motion.
(a) Terminal Speed:
The gravitational force acting on the skier can be calculated using the formula:
F_gravity = m * g
where m is the mass of the skier and equipment (m = 85.0 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Next, we need to calculate the drag force. The drag force acting on the skier can be calculated using the formula:
F_drag = 0.5 * C * ρ * A * v^2
where C is the drag coefficient (C = 0.150), ρ is the air density (ρ = 1.20 kg/m^3), A is the cross-sectional area of the skier (A = 1.30 m^2), and v is the velocity of the skier.
The frictional force can be calculated using the formula:
F_friction = μk * m * g
where μk is the coefficient of kinetic friction (μk = 0.0400).
At terminal speed, the drag force and the frictional force will balance the gravitational force. So we can write the equation:
F_drag + F_friction = F_gravity
0.5 * C * ρ * A * v^2 + μk * m * g = m * g
Simplifying and rearranging, we get:
0.5 * C * ρ * A * v^2 = (m * g) - (μk * m * g)
0.5 * C * ρ * A * v^2 = m * g * (1 - μk)
Dividing both sides by (0.5 * C * ρ * A), we get:
v^2 = (2 * m * g * (1 - μk)) / (C * ρ * A)
Taking the square root of both sides, we get:
v = sqrt((2 * m * g * (1 - μk)) / (C * ρ * A))
Let's substitute the given values and calculate the terminal speed:
v = sqrt((2 * 85.0 kg * 9.8 m/s^2 * (1 - 0.0400)) / (0.150 * 1.20 kg/m^3 * 1.30 m^2))
v ≈ 58.77 m/s
Therefore, the terminal speed is approximately 58.77 m/s.
(b) Variation in Terminal Speed:
To find the corresponding variation in the terminal speed when the drag coefficient changes slightly (dC), we need to differentiate the terminal speed equation with respect to C:
d(v) / d(C) = d / d(C) [sqrt((2 * m * g * (1 - μk)) / (C * ρ * A))]
Using the quotient rule of differentiation, we get:
d(v) / d(C) = -sqrt((2 * m * g * (1 - μk)) / (C^3 * ρ * A))
Let's substitute the given values into the equation:
d(v) / d(C) = - sqrt((2 * 85.0 kg * 9.8 m/s^2 * (1 - 0.0400)) / ((0.150)^3 * 1.20 kg/m^3 * 1.30 m^2))
d(v) / d(C) ≈ -1.01 m/s
Therefore, the corresponding variation in the terminal speed when the drag coefficient changes slightly is approximately -1.01 m/s.
To find the terminal speed of the skier, we can consider the forces acting on them. The two main forces are the drag force and the frictional force.
(a) Terminal speed is the speed at which the net force on an object becomes zero. At terminal speed, the drag force equals the frictional force, and there is no further acceleration.
1. Drag force: The drag force acting on the skier can be calculated using the equation:
Fd = 0.5 * ρ * A * C * (v^2)
where Fd is the drag force, ρ is the air density, A is the cross-sectional area of the skier, C is the drag coefficient, and v is the velocity of the skier.
2. Frictional force: The frictional force acting on the skier can be calculated using the equation:
Ff = μk * m * g
where Ff is the frictional force, μk is the coefficient of kinetic friction, m is the mass of the skier, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
At terminal speed, Fd = Ff, so we can equate the two equations:
0.5 * ρ * A * C * (v^2) = μk * m * g
Now, we can solve for the terminal speed (v).
1. Rearrange the equation:
v^2 = (2 * μk * m * g) / (ρ * A * C)
2. Take the square root of both sides:
v = √[(2 * μk * m * g) / (ρ * A * C)]
3. Substitute the given values into the equation:
v = √[(2 * 0.0400 * 85.0 * 9.8) / (1.20 * 1.30 * 0.150)]
v ≈ 37.7 m/s
Therefore, the terminal speed of the skier is approximately 37.7 m/s.
(b) To find the corresponding variation in terminal speed as a result of a slight variation 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)]
Now, we can substitute the given values into the equation:
dv/dC = -√[(2 * 0.0400 * 85.0 * 9.8) / (1.20 * 1.30 * 0.150^3)]
dv/dC ≈ -24.7 m/s
Therefore, the corresponding variation in terminal speed for a slight change in the drag coefficient would be approximately -24.7 m/s.