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.
first
Air friction drag = (1/2} rho v^2 A = 0.6*v^2*1.30 = 0.78 v^2
then do problem (If v is constant solve for v)
gravity force down slope = m g sin 39.5
forces up slope = 0.78 v^2 + 0.04*85*9.81* cos 39.5
if a = 0, down slope force = up slope force.
.78 v^2 + 0.04*85*9.81* cos 39.5 = 85*9.81*sin39.5
To find the acceleration of the skier, we need to calculate the net force acting on the skier. The net force can be found by subtracting the forces that the skier from the forces that propel the skier.
The forces that the skier are the air drag force and the kinetic frictional force.
1. Air drag force:
The air drag force can be calculated using the equation:
Fdrag = 0.5 * ρ * C * A * v^2
where:
- Fdrag is the drag force
- ρ is the air density (1.20 kg/m^3)
- C is the drag coefficient (0.150)
- A is the cross-sectional area of the skier (1.30 m^2)
- v is the speed of the skier
2. Kinetic frictional force:
The kinetic frictional force can be calculated using the equation:
Ffriction = μk * m * g
where:
- Ffriction is the frictional force
- μk is the coefficient of kinetic friction (0.0400)
- m is the mass of the skier and equipment (85.0 kg)
- g is the acceleration due to gravity (9.8 m/s^2)
Now, we can calculate the net force and then the acceleration:
Net Force = Fdrag + Ffriction
Substituting the equations for Fdrag and Ffriction, we get:
Net Force = 0.5 * ρ * C * A * v^2 + μk * m * g
Finally, we can find the acceleration (a) using Newton's second law:
a = Net Force / m
By substituting the values given in the problem, we can calculate the acceleration of the skier.