How to calculate the kinematics of an object pushed up an inclinex plane?

[ac7597]

Homework Statement

Joe puts on his skis and heads for the slopes. A fresh layer of new snow has fallen, making the coefficient of kinetic friction between his skies and the snow only μk=0.037. He heads for the Starlight Run, which features a long slope at a constant angle of θ=17.8 degrees above the horizontal.
The resort has a new device to bring skiers to the top of the run: instead of a chair lift or a moving rope, a large cylinder sits at the base of the hill. Skiers crouch down and slide into the cylinder, then press a button. Compressed gas shoots the skier out of the cylinder with an initial speed v=30.3 m/s, and they slide up the hill.

How far up along the slope will Joe slide before coming to a momentary halt?
How long will it take him to reach the peak of his motion?
After coming to a momentary halt, Joe starts to slide down the hill. How long will it take him to reach the bottom?

Relevant Equations

g=9.8 m/s^2

Homework Statement: Joe puts on his skis and heads for the slopes. A fresh layer of new snow has fallen, making the coefficient of kinetic friction between his skies and the snow only μk=0.037. He heads for the Starlight Run, which features a long slope at a constant angle of θ=17.8 degrees above the horizontal.
The resort has a new device to bring skiers to the top of the run: instead of a chair lift or a moving rope, a large cylinder sits at the base of the hill. Skiers crouch down and slide into the cylinder, then press a button. Compressed gas shoots the skier out of the cylinder with an initial speed v=30.3 m/s, and they slide up the hill.

How far up along the slope will Joe slide before coming to a momentary halt?
How long will it take him to reach the peak of his motion?
After coming to a momentary halt, Joe starts to slide down the hill. How long will it take him to reach the bottom?
Homework Equations: g=9.8 m/s^2

I made a component diagram of the forces acting on the skier. I got a horizontal force equation as m(ax) = F - u(cos17.8)mg-sin(17.8)mg, but am confused on how to proceed forward.

 

[Ssnow]

Hi, to start I suggest to use the energy considering also the work done by the friction ...
Ssnow

 

[ac7597]

I included frictions as Ff=u(Fn)

 

[ac7597]

Hello,
I've apparently misused the initial velocity : I was supposed to use 30.3m/s instead cos(17.8)*30.3. I rework the force component, and this is my answer.

 

[Ssnow]

(Sure you must use the decomposition on the incline plane ...). You can start to fix the variable $h$ that is final height (what you must find...). Now at the beginning you have a kinetic energy $K$ that is transformed in potential energy $U$. The work $W$ done by the friction is equal to the difference $K-U$. So you have the equation $K-U=W$...
Ssnow

 

[Ssnow]

Remember that the space covered by the skier along the inclinate $x$ plane is related to the maximal height $h$ by the relation $h=x\cdot \sin{\theta}$...
Ssnow

 

[Delta2]

Hello
Though the images don't look good (not good handwriting either) and I had to use software to rotate and magnify them (ok that was not hard but still) I recommend you learn to write equations with $\text LATEX$ commands. It will require for you some extra time to learn it and type your work every time you have a problem , but I am sure it will be beneficial at the end.

I found your work and equations correct though I didn't check the arithmetic.

Also i believe it would be beneficial to do it using the energy approach like @Ssnow suggests so you can cross check the results.

 

[ac7597]

Thanks