Energy conservation (potential, kinetic, spring)

[jegues]

Hello again,

The question and free body diagram as well as my attempt at the solution are all in the attached photo.

Thanks in advance!

 

[jegues]

My last thread was moved, I'm assuming I posted this one in the wrong place as well, sorry!

 

[nrqed]

Hello again,

The question and free body diagram as well as my attempt at the solution are all in the attached photo.

Thanks in advance!

Watch out. If you use x for the vertical distance shown in your figure, the spring is not compressed a distance x! Do you see why?

 

[jegues]

The variable I should be using in the calculation for the Usf (final spring energy) should be r, where r is defined as follows, r = x/sin(theta) correct?

 

[nrqed]

The variable I should be using in the calculation for the Usf (final spring energy) should be r, where r is defined as follows, r = x/sin(theta) correct?

That's it!

 

[jegues]

Thanks again friend, I seem to only need the slightest push in order to figure out where I went wrong.

 

[nrqed]

Thanks again friend, I seem to only need the slightest push in order to figure out where I went wrong.

You're welcome. That's a good sign, you know? Because it means that you just need to see the little tricks to complete the problem and once you know the trick, you are all set for other problems. Some people need help with all the steps, from the very beginning. So I would say that you are doing very well!

 

[jegues]

Hmmm... Apparently I still can't get to the correct answer.

I'm using the equation:

mgh + mgx - 1/2k(r)^2 = 0; Giving me,

-16(x^2) + 0.784x + 0.4704 = 0; This doesn't seem to give me the correct zeros?

 

[Yuqing]

The answer that equation gives you is the vertical compression of the spring. You need to find the compression along the slope of the incline.

 

[Redbelly98]

My last thread was moved, I'm assuming I posted this one in the wrong place as well, sorry!

FYI for in future: from the Physics Forums main page, look for "Homework and Coursework Questions" and click there.