an extreme skier starting from rest, coasts down a mountain that makes an angle 25.0¨¬with the horizontal. The coeffiecient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 8.4m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 2.80m below the edge. How fast is she going just before she lands?
Frictional force, F
= μmg cos(θ)
Work done to overcome kinetic friction
= F*D
= 8.4F
Difference in elevation, H
= 8.4 sin(&theta)+2.8 m
Energy left (in the form of KE)
= mgH-8.4F
Equate Kinetic energy with potential energy
(1/2)mv² = mgH-8.4(μmg cos(θ))
All quantities are known except v.
Solve for v (9.7 m/s)
Check my thinking and calculations.
To find the speed at which the skier lands, we need to analyze the motion in two parts:
1. Coasting down the mountain:
Let's calculate the acceleration of the skier while coasting down the mountain using the force equation:
Frictional Force = coefficient of kinetic friction * normal force
The normal force is equal to the gravitational force acting on the skier, which can be calculated using the skier's weight:
Weight = mass * acceleration due to gravity
Using these equations, we have:
Frictional Force = coefficient of kinetic friction * weight
Since the frictional force is the force that causes the acceleration, we can set it equal to the mass multiplied by the acceleration:
Frictional Force = mass * acceleration
Setting both sides equal, we get:
coefficient of kinetic friction * weight = mass * acceleration
Simplifying, we have:
coefficient of kinetic friction * mass * acceleration due to gravity = mass * acceleration
Dividing both sides by mass and simplifying further, we find:
coefficient of kinetic friction * acceleration due to gravity = acceleration
Now, we can use trigonometry to calculate the acceleration along the incline:
Acceleration = gravitational acceleration * sin(angle of incline)
Using the given values:
acceleration due to gravity = 9.8 m/s^2
angle of incline = 25.0°
Plugging in these values, we have:
Acceleration = 9.8 m/s^2 * sin(25.0°)
2. Falling off the cliff:
Since the skier is not acted upon by any horizontal forces after leaving the edge of the cliff, the only force acting on the skier is gravity. Therefore, the only acceleration acting on the skier is the acceleration due to gravity, which is 9.8 m/s^2.
Now, we can calculate the final speed of the skier just before she lands using the equations of motion.
Let's assume the initial velocity is 0 m/s. The vertical distance traveled is given as 2.80 m, and the acceleration is -9.8 m/s^2 (negative because the acceleration is opposite to the direction of motion).
The equation we can use to solve for the final velocity is:
Final Velocity^2 = Initial Velocity^2 + 2 * acceleration * distance
Substituting the given values, we get:
Final Velocity^2 = 0^2 + 2 * (-9.8 m/s^2) * (-2.8 m)
Simplifying, we have:
Final Velocity^2 = 54.56 m^2/s^2
Taking the square root of both sides, we find:
Final Velocity = √(54.56 m^2/s^2)
Evaluating this, we get:
Final Velocity = 7.39 m/s
Therefore, the skier is going approximately 7.39 m/s just before she lands.
To find the speed of the skier just before she lands, we need to analyze the forces acting on her during two different stages: while coasting downhill and during free fall after leaving the cliff.
First, let's determine the speed of the skier when she reaches the edge of the cliff.
1. Coasting Downhill:
When the skier is coasting downhill, the forces acting on her are the gravitational force (mg), the normal force (N), and the frictional force (f).
The gravitational force can be split into two components: parallel to the slope (mg * sinθ) and perpendicular to the slope (mg * cosθ), where θ is the angle of the slope.
The equation of motion in the horizontal direction is:
f = μN
Substituting the equation for frictional force (f) with the given coefficient of kinetic friction (μ) and the normal force (N = mg * cosθ), we get:
μ * N = μ * mg * cosθ
The equation of motion in the vertical direction is:
mg * sinθ = N + mg * cosθ
We need to solve these two equations together to find the acceleration (a) and the speed (v) of the skier when she reaches the edge of the cliff.
Using the equation of motion in the horizontal direction, we can rewrite it as:
μ * mg * cosθ = ma
Simplifying the equation of motion in the vertical direction, we get:
mg * sinθ = mg * cosθ + ma
=> mg (sinθ - cosθ) = ma
=> (sinθ - cosθ) = a
Since a = μ * g * cosθ, we can substitute this value into the previous equation:
(sinθ - cosθ) = μ * g * cosθ
Now we can solve for the acceleration (a) using the given angle of the slope (θ):
a = (sinθ - cosθ) / (1 + μ * cosθ)
Next, we can calculate the speed (v) of the skier using the equation of motion:
v^2 = u^2 + 2as
Since the skier starts from rest, the initial velocity (u) is 0, and the distance (s) is the given distance before the cliff (8.4m). Plugging in these values, we find:
v^2 = 0 + 2 * a * 8.4
Now we have the speed (v) of the skier just before she reaches the edge of the cliff.
2. Free Fall After Leaving the Cliff:
When the skier leaves the cliff, she enters free fall. The only force acting on her is the gravitational force (mg) pulling her downwards.
The equation of motion in the vertical direction is:
mg - mg = ma
0 = ma
Since the net force is zero, the acceleration (a) is also zero during free fall.
Now we can find the speed (v') of the skier when she lands at a point 2.80m below the edge.
Using the equation of motion:
v'^2 = v^2 + 2as
Since the acceleration (a) is zero during free fall, the equation simplifies to:
v'^2 = v^2 + 2 * 0 * 2.80
v'^2 = v^2
This means the speed (v') of the skier when she lands is equal to the speed (v) she had just before reaching the edge of the cliff.
Therefore, the speed of the skier just before she lands is equal to the speed calculated earlier using the given distance (8.4m) before the cliff.