How Fast Did Klaus Spinka Accelerate on Grass Skis?

[OttoVon]

Homework Statement

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?
The answer is:
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

Homework Equations

Friction=Mn
Efx:
Efy:

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

 

[haruspex]

What is the value of a_y?

 

[OttoVon]

What is the value of a_y?

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

 

[haruspex]

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

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?

 

[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?

 

[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?

 

[haruspex]

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?

I asked about the velocity perpendicular to the slope.

 

[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.

 

[haruspex]

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.

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

 

[Chestermiller]

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

 

[dean barry]

Can i go back to finding the acceleration ?
g I've 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

 

[haruspex]

Can i go back to finding the acceleration ?
g I've 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

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.

 

[Chestermiller]

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.

I sent an official Warning to Dean Barry about this earlier today.

Chet