Dynamics Skier on a Slope Problem

[IbrahimA]

Homework Statement

A skier on a slope inclined at 4.7 degrees to the horizontal pushes on ski poles and starts down the slope. The initial speed is 2.7 m/s. The coefficient of kinetic friction between skis and snow is 0.11. Determine how far the skier will slide before coming to rest.

Answer on the book: 13m

Homework Equations

Fnetx = (-Ff) + Fa + (-Fgx)
Fnety = Fn + (-Fgy)v2^2 = v1^2 + 2ad

The Attempt at a Solution

I went by creating two equations:

Fnetx = (-Ff) + Fa + (-Fgx)
ma = -Fn(u) + Fa + mg(sin4.7)
m was canceled out on both sides.
Fn(u) = Fa + g(sin4.7) - a
(1) Fn = Fa + g(sin4.7) - a / u

Fnety = Fn + (-Fgy)
ma = Fn + (m)(-g)(cos4.7)
again canceled out m on both sides
(2) Fn = a + gcos4.7

Correction: Equation 2 is Fn = gcos4.7

And so I did substitution and subbed in both equations to solve for acceleration (a) but the problem is that Fa is also unknown:

(9.8)cos4.7 = Fa + (9.8)(sin4.7) - a / 0.11

So I'm wondering if there is a better way to do this problem, because I feel like I'm doing to wrong.

I included a fbd diagram:

 

[haruspex]

In your equation (2), what acceleration does a represent? Acceleration into the snow?

 

[IbrahimA]

In your equation (2), what acceleration does a represent? Acceleration into the snow?

Looking back at equation 2, I feel like the net force on the y-axis is 0. Because from my understanding, the skier is only moving on the x-axis being on the incline, where the forces on the y-axis are cancelling each other out.
Correction of equation 2: Fn = gcos4.7
But even then that still leaves two unknowns, both Fa and a.

 

[haruspex]

But even then that still leaves two unknowns, both Fa and a.

And what applied force does Fa represent?

 

[IbrahimA]

And what applied force does Fa represent?

Fa is the applied force that the skier is pushing on the ski poles thus moving up the incline.

 

[haruspex]

Fa is the applied force that the skier is pushing on the ski poles thus moving up the incline.

The skier is moving down the incline and only uses the poles to get going. Having reached 2.7m/s, the poles are no longer used, and the skier slides to a stop.

 

[IbrahimA]

The skier is moving down the incline and only uses the poles to get going. Having reached 2.7m/s, the poles are no longer used, and the skier slides to a stop.

I have redone my equations, with:
Equation 1 : Fn = a + gsin4.7 / u
Equation 2 : Fn = mgcos4.7

But again same situation, there are still two unknowns being m and a, since I can't cancel out m on the y-axis as shown below:
Fnety = Fn + (-Fgy)
0 = Fn + (m)(-g)(cos47)
Fn = mgcos4.7

 

[IbrahimA]

The skier is moving down the incline and only uses the poles to get going. Having reached 2.7m/s, the poles are no longer used, and the skier slides to a stop.

Algebra Question
In the equation as shown below:
(m)(a) = (u)(g)(m)(cos4.7) + (m)(-g)(sin4.7)
Can I cancel out all the m's on both sides and result into this?:
a = (u)(g)(cos4.7)+(-g)(sin4.7)
Or is that not possible?

 

[haruspex]

Algebra Question
In the equation as shown below:
(m)(a) = (u)(g)(m)(cos4.7) + (m)(-g)(sin4.7)
Can I cancel out all the m's on both sides and result into this?:
a = (u)(g)(cos4.7)+(-g)(sin4.7)
Or is that not possible?

Of course you can cancel the m's. All you need to assume is that m is not zero, and you can safely do that.

 

[IbrahimA]

Of course you can cancel the m's. All you need to assume is that m is not zero, and you can safely do that.

Oh okay, now it makes sense. Thank you.