Translation Motion Incline problem

[TheJourney]

What is the acceleration of an 80-kg skier going down the track if θ = 45°?

A. 6.9 m/s^2
B. 9.8 m/s^2
C. 13.9 m/s^2
D. 80 m/s^2

What seems to be tricking me up is that I don't know where a "acceleration" should go.
So if the skier is on an incline the force of gravity will be acting on it equivalent to mg. I also know the skier will be moving parallel to the incline plane with a normal force equivalent to mgsin45. I substitute mgsin45 for masin45 where a is the acceleration and setting the 2 equations equal to each other I get mg = masin45.
Solving for a equals approx 14, answer choice C. How do I know that equation should be rewritten as ma = mgsin45 instead in order to yield the correct answer A?

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[Teleologist]

Draw a free body diagram.

 

[TheJourney]

I did, I might be doing it incorrectly.
Pointing down is the force of gravity which equal to mg... pointing parallel to incline plane is the normal force equal to mgsintheta.
From here I'm unsure where to put the acceleration, so I placed it in the mgsintheta equation.

 

[SuperSneaky24]

I hate these questions. In a college exam, I'd have time to draw the free body diagram, but on the MCAT you have so little time and quickly assume the wrong diagram. Basically you are drawing the free body diagram incorrectly. Think about your answer of 14m/s^2. There is no x-acceleration involved, yet somehow the skiier is still accelerating faster than gravity?

I drew the two diagrams to help you understand better.

EDIT: Whoops I meant to write mg/sin(theta) on the left side instead of sin(theta)/mg

 

[TheJourney]

I hate these questions. In a college exam, I'd have time to draw the free body diagram, but on the MCAT you have so little time and quickly assume the wrong diagram. Basically you are drawing the free body diagram incorrectly. Think about your answer of 14m/s^2. There is no x-acceleration involved, yet somehow the skiier is still accelerating faster than gravity?

I drew the two diagrams to help you understand better.

EDIT: Whoops I meant to write mg/sin(theta) on the left side instead of sin(theta)/mg

Thanks for your response Super! You're right, I just looked back at my drawings and I was estimating the angle from the wrong place. However, I'm still not sure how to differentiate which acceleration to solve for. For example if we left them both as "g" mg = mgsintheta, how would you know you have to solve for the one on the left hand side or right hand side?

 

[SuperSneaky24]

Yup, no prob! As for your next question, well if you look at the question first glance, you see that they're asking for acceleration only. As mass is independent of acceleration, we shouldn't even worry about the mass of the skier anyways.

You'll end up just getting a = gsin(theta)

However, if we do it do the force way like you did, we aren't supposed to use g on the left side. We definitely use g on the right side, as that deals with the acceleration from gravity. The left side should be m*a, which is the force felt of the skier on the slope. In this case, the a on the left side would represent the acceleration that the skier feels, while the g on the right side is the acceleration due to gravity. I'm sure if you do this problem with the new free-body diagram again, you'll understand.

 

[TheJourney]

Ahhhh, gotcha!! Thank you so much for clearing that up! And ya I'll def try a bunch of other incline problems using the right angle this time haha.