/sci/ - Science & Math » Thread #11711415

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fuck this fucking fuck

Anonymous Mon 25 May 21:51:11 2020 No.11711415 View Reply Original Report

Quoted By: >>11711717 >>11711792

I've been stuck on this forever and can't figure it out, I know I didn't include the initial velocity but my brain is to mush to do it and it's due soon. Could I get some help please.

Anonymous

Anonymous Mon 25 May 2020 21:53:12 No.11711419 Report

Quoted By: >>11711712 >>11711837

Everything is an FBD

Anonymous

Anonymous Mon 25 May 2020 23:42:47 No.11711712 Report

Quoted By:

>>11711419
I know, but I keep getting the wrong answer when trying to add initial velocity.

Anonymous

Anonymous Mon 25 May 2020 23:44:58 No.11711717 Report

Quoted By:

>>11711415
do you know how to use conservation laws? if so there you go.

Anonymous

Anonymous Tue 26 May 2020 00:03:06 No.11711756 Report

Quoted By: >>11711920

You don’t even have to get up and go into your TAs office anymore you lazy fuck. Call him on zoom and get the answer

Anonymous

Anonymous Tue 26 May 2020 00:16:17 No.11711792 Report

Quoted By:

>>11711415
Do your own homework, slowly and carefully and don't put in any fuckin numbers until the end, ever

Anonymous

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Anonymous Tue 26 May 2020 00:34:41 No.11711837 Report

Quoted By: >>11711913 >>11712037

>>11711419
Total energy at the start is (mv^2)/2. Total energy at the bottom is (kinetic energy at the bottom, 0)+(gravitational potential energy lossed)+(potential energy stored in spring). Equate these two things and rearrange to get $\frac{1}{2}kx^2-mg\sin\theta x-mg\sin\theta d-\frac{1}{2}mv^2=0$
so you literally apply the quadratic formula.

Anonymous

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Anonymous Tue 26 May 2020 00:55:46 No.11711913 Report

Quoted By: >>11712037

>>11711837
Thanks but I'm doing something wrong.

Anonymous

Anonymous Tue 26 May 2020 00:58:06 No.11711920 Report

Quoted By:

>>11711756
It literally takes a 9 days for the TA to get back to me and the answer is always fucking useless. I don't usually do this, I have a 4.2/4.3 GPA its just that I'm going crazy over this one.

Anonymous

Anonymous Tue 26 May 2020 01:08:27 No.11711940 Report

Quoted By: >>11711948 >>11712008

Let x be the displacement of the spring from equilibrium when the mass is stationary. Initially it has a kinetic energy and gravitational potential. At the bottom it has no velocity so its kinetic energy is 0 at that time.

$E_i = T_i + V_i = \frac{1}{2} m v^2 + mgd \sin{\theta}\\<br />E_f = T_f + V_f = 0 + \frac{1}{2} k x^2\\<br />$

Since we have energy conservation the difference is 0;

$E_f - E_i = \frac{1}{2} k x^2 - (\frac{1}{2} m v^2 + mgd \sin{\theta})\\<br />x = \frac{\pm \sqrt{4(\frac{1}{2}k)(\frac{1}{2} m v^2 + mgd \sin{\theta})}}{k}$

You can finish the rest for yourself.

Anonymous

Anonymous Tue 26 May 2020 01:11:32 No.11711948 Report

Quoted By: >>11712008

>>11711940
nevermind this, ignore, I forgot the (d-x) it has to travel

Anonymous

Anonymous Tue 26 May 2020 01:30:45 No.11712008 Report

Quoted By:

>>11711940
>>11711948
Thanks anyways

Anonymous

Anonymous Tue 26 May 2020 01:45:36 No.11712037 Report

Quoted By:

>>11711837
>>11711913
Never mind I see where I went wrong thank you very much!

Anonymous

Anonymous Tue 26 May 2020 02:07:44 No.11712071 Report

Quoted By:

simple; find the lagrangian, use the euler-lagrange equation and get x.

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