Recent content by Dark Visitor

SERIOUSLY! SWEET! That just made my night! Thanks a lot man! I PASSED PHYSICS for the semester! :biggrin:

I can't remember exactly how I did it (it was due today, so I did what I could last night and turned it in.) I think I found Wf to be -4000, which my final answer turned out to be 4550 J, which was an answer, so I left it at that. Thanks for your help though. You got me that far.

Friction should oppose the motion of the wheelbarrow, making it negative. I also worked out the numbers, which I will show: Ki = 0 Ui = (500 N)(17.101) = 8550.5036 Uf = (500 N)(0) = 0Now doesn't that mean that Kf = Ui + Wf?

K = 1/2(m)(v)2 Umg = mgy (y is vertical height above/below y=0 position)

Then what is the angle? That confuses me.

So is what I just posted correct? And how do I find each individual piece to your equation?

Well, trying it out, I got: Wby F = F*\Deltax*sin\theta = (80 N)((50 m)(sin20)) = 1368.0806 Is that right? And if so, what do I need to find next?

Okay, I think I can get that one eventually. There is a guy on there helping me now, so thanks to both of you for your help. I know it was a pain in the butt. But I appreciate it. :approve:

So are you saying that for this part, all I am focusing on is the work done by friction? Wnc = Wf

Okay, I redid it using the new equation. I think I got the same things you did, which is good. :) But my question is: are we doubling it for the same reason we did before? (A = 2x)

Why thanks? And why would we use that one now?

Why is 2\pi in the numerator? That's not the equation I had written down... I do have another one that says: Tspring/mass = 2\pi/\sqrt{}k/m

So what do I do? I can't find where my error is...

I don't know. I got 3.77 m, which converts to 377, or 380, cm. But my answer is 38 cm. I don't see what I did wrong...

I think I understand. My only issue now is, when I got 3.77 m as my A, how is the answer 38 cm? Did I mess up somewhere with my math?