# Conservation of Mechanical Energy and Kinetic Friction

## 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.

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.

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

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)

Last edited:
BvU