Speed skier travelling up a hill

[Ly444999]

Homework Statement

A speed skier is traveling horizontally at a constant speed of 5.80 m/s when she approaches a snow-covered hill that has a slope of 23.0° above the horizontal. The coefficient of kinetic friction between the skis and the hill is 0.350 and the combined mass of her and her skis is 77.0 kg. If she decided to glide up the hill, how far would she make it before she comes to a complete stop?

Homework Equations

Fnet = m*a
Fk = μ Fn
Fg = m(-g)
Fgx = m(-g)sinθ
Fgy = m(-g)cosθ

The Attempt at a Solution

So first I solved for the y component of the force due to gravity:
Fgx = (77.0kg)(-9.8m/s²)*cos23 = -694.61296

Then I solved for the kinetic friction with the the y component of the gravity force equaling the normal force:
Fk = (0.350)(-694.61296) = -243.11454

Then I went on to solve for acceleration using the Fnet formula with the x component of gravity subtracting the kinetic friction force:

m*a = Fgx - Fk
(77.0kg)*a = [(77.0kg)(-9.8m/s²)*sin23] - 243.11454
a = -6.98650

I was just wondering is this right so far?

 

[haruspex]

is this right so far?

Looks right. (With that coefficient of friction for skis on snow, the skiing industry would collapse.)

 

[Ly444999]

Looks right. (With that coefficient of friction for skis on snow, the skiing industry would collapse.)

So I am supposed to use negative 9.8 (or whatever decimal places I use), depending on whether I make up or down positive (I made down negative), on both x and y component?
Also is that Fnet equation right Fgx subtract Fk? I'm having a lot of doubts and confusions in my calculations from which value should be positive or negative.

 

[haruspex]

So I am supposed to use negative 9.8 (or whatever decimal places I use), depending on whether I make up or down positive (I made down negative), on both x and y component?
Also is that Fnet equation right Fgx subtract Fk? I'm having a lot of doubts and confusions in my calculations from which value should be positive or negative.

Clearly both the friction and the downslope component of gravity will contribute to the slowing of the skier, so the important thing is that they should have the same sign. The other main opportunity for errors is in confusion between sine and cosine, but I confirm you have that right too.

 

[Ly444999]

I plugged in that acceleration into V_f² = V_i² +2*a*d and I ended up getting 4.81500 m and that answer was wrong apparently did I do something wrong?
Final velocity I'm almost certain is 0.
Initial velocity was given at 5.80 m/s
acceleration I solved to be -6.98650

Edit: Oops I forgot to multiply the acceleration by 2 before I divided the initial velocity by that

 

[haruspex]

I plugged in that acceleration into V_f² = V_i² +2*a*d and I ended up getting 4.81500 m and that answer was wrong apparently did I do something wrong?
Final velocity I'm almost certain is 0.
Initial velocity was given at 5.80 m/s
acceleration I solved to be -6.98650

I get about half that. Did you forget the 2 in 2ad?

 

[Ly444999]

I get about half that. Did you forget the 2 in 2ad?

Yeah I did ooops lol thanks for quick reply though.

 

[haruspex]

Yeah I did ooops lol thanks for quick reply though.

Ok. Fwiw, a typical coefficient for ski on snow would be less than 0.1.