# Homework Help: Skier Sliding Down Ski Slope

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1. Oct 25, 2014

### OttoVon

1. The problem statement, all variables and given/known data
The record speed for grass skiing was set in 1985 by Klaus Spinka, of
Austria. Suppose it took Spinka 6.60 s to reach his top speed after he
started from rest down a slope with a 34.0° incline. If the coefficient of
kinetic friction between the skis and the grass was 0.198, what was the
magnitude of Spinka’s net acceleration? What was his speed after 6.60 s?
a=3.87m/s^2
For the speed after 6.60s
velocity final=25.5m/s^2

What we Have
Time: 6.60s
Degree: 34
mg:?
n:?
M:0.198
From rest: V initial = 0 m/s
a=?
Speed after: 6.60s
2. Relevant equations
Friction=Mn
Efx:
Efy:

3. The attempt at a solution
This is what I did so far:
EFx: Friction-mgsin(34)=max
EFy:Normal Force-mgcos(34)=may
Added the mg's to the other side
Friction=mgsin(34)+max
Normal Force=mgcos(34)+may

Then plugged into the equation to find friction friction=(0.198)(normal force)
mgsin34+max=(mgcos34+may)(0.198)
The problem is that everything will cancel out so that is preventing me from finding acceleration

2. Oct 25, 2014

### haruspex

What is the value of ay?

3. Oct 26, 2014

### OttoVon

What do you mean? It didn`t specify a value is it zero?

4. Oct 26, 2014

### haruspex

It represents the acceleration perpendicular to the slope, right? Does the skier stay in contact with the slope? So what's the velocity perpendicular to the slope? Is it constant?

5. Oct 26, 2014

### OttoVon

He stays in contact with the slope, the velocity wouldn't be constant wouldn't it because acceleration is there to make it increase/decrease right?

6. Oct 26, 2014

### billy_joule

If the skier stays in contact with the slope what is his displacement in the y direction? What does that displacement say about the velocity and acceleration in that direction?

7. Oct 26, 2014

### haruspex

8. Oct 27, 2014

### dean barry

Your initial acceleration looks correct, you have assumed contant acceleration up to the time limit, whereas with air drag involved the acceleration rate would diminish during the journey, the terminal velocity never being reached.
This involves maths of a greater sophistication than simply newtons rules.

9. Oct 27, 2014

### haruspex

I don't believe drag is supposed to be taken into account here. It would leave insufficient information.

10. Oct 27, 2014

### Staff: Mentor

What these guys are asking is "If the skis are always in contact with the slope, what is their velocity component perpendicular to the slope?"

Chet

11. Oct 28, 2014

### dean barry

Can i go back to finding the acceleration ?
g ive asumed at 9.81 m/s/s
The root equation is :
acceleration = f / m
But :
f = ( m * g * sine 34 ° ) - ( m * g * cosine 34 ° )
f = ( 5.4856 m ) - ( 1.6103 m )
Subtracting, you get :
f = 3.8753 m
Back to the root equation :
acceleration = ( 3.8753 m ) / m
The m's cancel, you get :
acceleration = 3.87538 m/s/s

If you then apply this acceleration constantly for 6.6 seconds youre velocity will be 25.58 m/s

12. Oct 28, 2014

### haruspex

You should know by now that you should not post complete solutions until the original poster has posted a solution. Until then, just helpful hints please.

13. Oct 28, 2014

Chet