Calculating Acceleration of a Skier on an Inclined Slope

[Mirole]

Homework Statement

You, a 75-kg skier, glide straight down a snow-covered slope inclined at 15 degrees to the horizontal over spring break. Let's be realistic, what is your acceleration(magnitude and direction) Assume your skis are wood and the snow is dry. mu_k on snow is 0.060.

Homework Equations

Fnet=ma

The Attempt at a Solution

Fx = -fk+mgsin\theta = ma
Fy = n = mgcos\theta

-mu_k*mgcos\theta = ma - mgsin\theta
a = -mu_k*g*cos\theta + gsin\theta
a = (.060)(9.8)cos15 + 9.8sin15
a = 8.22 m/s²

Did I do the first part right? And, how would I find the direction?

 

[Delphi51]

a = -mu_k*g*cosθ + gsinθ
a = (.060)(9.8)cos15 + 9.8sin15

Looks like you dropped a minus sign in that step.
The friction force opposes the parallel component of gravity, so there ought to be subtraction rather than addition.

 

[Mirole]

Wow, I did that by mistake and didn't even catch it. Thanks, so would 1.97m/s² be correct?