Conservation of Mechanical Energy and Kinetic Friction

[martepartay]

Homework Statement

Question in entirety: Vladimir Putin has contacted you to redesign the ski jump at the Olympic games. In the ski jump, contestants begin at rest at the starting gate on top of a hill inclined at an angle of 30 with respect to horizontal. They then accelerate down the hill, before launching off a ramp at the end of the incline. Your task is to determine what height above the ramp the starting gate should be moved to, in order that the skiers launch with a speed of 35 m/s. Assume that the coefficient of kinetic friction between the skis and snow is 0.1, and that air resistance is negligible since the skiers will be wearing aerodynamic suits.

Homework Equations

E = PE + KE
PE = mgh
KE = 1/2mv^2
Fk = μN
ΔE = Wnc

The Attempt at a Solution

I have got an answer but I am nearly positive it is not correct. First I made a a diagram, though I don't have a scanner so I can't upload it unfortunately. To solve I deduced that mgh = 1/2mv^2 + Energy of Kinetic Friction since energy can not be created or destroyed. Then I got the equation mgh = 1/2mv^2 + μmgcos(θ) believing that mgcos(θ) must be the normal force which is where things get a bit less certain for me conceptually. I then, after cancelling mass and plugging in my values for v, g, μ, and θ, got 62.59m for h. The number seems plausible, however, I don't believe I have fully grasped how the kinetic friction comes into play. In checking my answer I predicted that v in a world without friction should be at least >35.5. However I got 35.027 when simply doing mgh=1/2mv^2. Thus, I think that my energy of kinetic friction is too low. Friction essentially didn't matter if this were true. In the energy of kinetic friction shouldn't displacement be involved? I'm also now unsure if I used the 30 angle correctly.

Thanks in advance physics community!

 

[BvU]

mgh = 1/2mv^2 + μmgcos(θ) is dimensionally incorrect! What dimension is missing (you kind of say this already) ?

 

[martepartay]

I knew it! I knew it had to be involved. The distance kinetic friction acts on must be factored in since W= force x distance. I'm a bit unsure what the distance is though. I comprised the equation mgh - 1/2mv^2 = FxD = μmgcos(θ) xD

Then I made a triangle with height h, base A, and hypotenuse. I figured the hypotenuse would be h/Sin(30) and plugged that into the equation thus making mgh-1/2mv^2 = μmgcos(30) X h/sin(30)

 

[BvU]

Looks good to me ! Greetings to Vlad!

And my compliments for your initial post: you check things and sense there is something amiss. Keep that up!

 

[HalfLostAndSearching]

Dear Vladimir Putin,

Thank you for reaching out to me for assistance in redesigning the ski jump at the Olympic games. After considering the given information, I have come up with a solution that may be of use to you.

Based on the conservation of mechanical energy, we can equate the initial potential energy to the final kinetic energy of the skier. Thus, we have the equation mgh = 1/2mv^2, where m is the mass of the skier, g is the acceleration due to gravity, h is the height of the starting gate, and v is the final velocity of the skier.

However, we must also take into account the kinetic friction between the skis and snow. The force of kinetic friction, Fk, is equal to the coefficient of kinetic friction, μ, multiplied by the normal force, N. In this case, the normal force is equal to the component of the weight of the skier that is perpendicular to the incline, which is mgcosθ. Therefore, we have Fk = μmgcosθ.

Using the work-energy theorem, we can relate the work done by kinetic friction to the change in mechanical energy of the skier. This can be expressed as ΔE = Wnc, where ΔE is the change in mechanical energy and Wnc is the work done by non-conservative forces, such as friction.

Substituting our equations for potential energy, kinetic energy, and work done by friction, we get mgh = 1/2mv^2 + μmgcosθ. Solving for h, we get h = (1/2v^2 + μgcosθ)^-1.

Plugging in the given values of v = 35 m/s, μ = 0.1, g = 9.8 m/s^2, and θ = 30°, we get h = 62.59 m as the height above the ramp that the starting gate should be moved to.

I hope this solution is helpful in your task of redesigning the ski jump. Please let me know if you have any further questions or concerns.

Best regards,