Stopping distance using velocity and friction coefficient on an incline plane

[Monic]

Homework Statement

A skier skiing downhill reaches the bottom of a hollow with a velocity of 20m/s, and then coasts up a hill with a 10° slope. If the coefficient of kinetic friction is 0.10, how far up the slope will she travel before she stops?

Homework Equations

aΔt=v2-v1
d= 1/2(v1 + v2)∆t
d=v1Δt + 1/2aΔt^2
v2^2=v1^1 + 2ad
F=ma
μ=Ff/Fn

The Attempt at a Solution

d=20^2 / 2(9.8)(sin10)(0.1)


I tried to substitute for a in a formuala I found online that worked previously for stopping distance on a horizontal plane.

 

[PBV]

First you have to take into account the force of gracity slowing her down on the slope as well. Which is F_grav= - mg sin(10).

Secondly the friction, which is the normal force times the constant of friction. That gives F_friction = - mg cos (10) * 0.1

The resulting force slowing her down would then be F = -mg ( 0.1cos(10) + sin(10) = ma, implying a = -g(0.1cos(10)+sin(10)).

Now using your top formula we get Δt = (v2-v1)/a = -20/-[g(0.1cos(10)+sin(10))].

Then using the second one gives:

d = 20^2 / 2[g(0.1cos(10)+sin(10))]

 

[Monic]

Thank you so much! That gets the exact answer in the book (75m) Much appreciated :)