Damon
answered
= (1/2)(1.2)(v^2)(1.8)(.4) = 0.432 v^2
friction drag = 0.06 (70*9.81)cos 30 = 35.7 Newtons
component of weight down slope = (70*9.81) sin 30 = 343 Newtons
so
343 = 35.7 + .432 v^2
solve for v
Explain Bot
answered
The drag force equation is given by:
Fdrag = 0.5 * C * P * A * V^2
Where:
- Fdrag is the drag force
- C is the drag coefficient
- P is the density of the fluid through which the object is moving (in this case, air)
- A is the cross-sectional area of the object
- V is the velocity of the object
Let's break down the values given in the question:
- Mass of skier (m) = 70 kg
- Height of skier (h) = 1.8 m
- Width of skier (w) = 0.4 m
- Friction (f) = 0.06
- Drag coefficient (C) = 1.1
- Density of air (P) = 1.2 kg/m³
First, we need to calculate the cross-sectional area (A) of the skier. Since the skier is upright and moving down the hill, the cross-sectional area is the product of the height and width of the skier:
A = h * w = 1.8 m * 0.4 m = 0.72 m²
Next, we need to calculate the gravitational force (Fg) acting on the skier. The force of gravity can be found using the formula:
Fg = m * g
Where:
- m is the mass of the skier (70 kg)
- g is the acceleration due to gravity (approximately 9.8 m/s²)
Fg = 70 kg * 9.8 m/s² = 686 N
Now, we can calculate the net force (Fnet) acting on the skier. The net force is the difference between the gravitational force and the frictional force:
Fnet = Fg - Ffriction
Ffriction = f * Fg
= 0.06 * 686 N
= 41.16 N
Fnet = 686 N - 41.16 N = 644.84 N
To find the terminal velocity, we set the net force equal to the drag force:
Fnet = Fdrag
644.84 N = 0.5 * C * P * A * V^2
Now we can solve for V, the terminal velocity of the skier.
V^2 = (2 * Fnet) / (C * P * A)
V = sqrt((2 * Fnet) / (C * P * A))
Substituting the values we have calculated:
V = sqrt((2 * 644.84 N) / (1.1 * 1.2 kg/m³ * 0.72 m²))
Calculating this expression will give you the terminal velocity of the skier going down the snow-covered hill.
