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Finding position wit the energy method

  1. Sep 26, 2009 #1
    Finding position with the energy method

    1. The problem statement, all variables and given/known data
    A block slides on a frictionless plane inclined at an angle [tex]\vartheta[/tex] above the horizontal, in the presence of a uniform gravitational field g. The block is realsed from rest at a distance L from the end of the plane. Use the energy method to derive an expression for the position x of the block on the plane as a function of time by evaluating t=[tex]\((dx/dt)^{-1}[/tex] dx


    2. Relevant equations



    3. The attempt at a solution

    Okay so I took the integral of t along starting from L-->0. My answer was t(L)=-(dtL/dx) which can be simplified to t(L)= -(1L/v). Am I correct in my thinking that the integral of t would give the position of the block as a function of t.
     
    Last edited: Sep 26, 2009
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  3. Sep 27, 2009 #2

    Redbelly98

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    Re: Finding position with the energy method

    Technically that should be

    dt = (dx/dt)-1dx​

    Guess that was simply a typo on your part.
    For this problem, you'll need an equation that relates to the energy of the block.

    The mistake here is in treating (dx/dt) as a constant when you do the integral. But it is not constant, the block is accelerating hence v changes.

    You'll need to write (dx/dt) as a function of x, and then do the integral. See my comment after "Relevant equations" above.
     
  4. Sep 27, 2009 #3
    yeah it was supposed to be dt, my mistake.

    okay well it took me a while (I had trouble with the integration) but I think I got the right answer, and I'll just upload an image instead because I'm not good with the P.F's latex formulas.

    http://img21.imageshack.us/img21/4167/cci2709200900000.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Sep 27, 2009 #4

    Redbelly98

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    You have the right idea, and are very close to the correct solution.

    Can you account for the slope of the plane in the expression for U(x)?
     
  6. Sep 27, 2009 #5
    sorry for the rather late reply but it took me a while to figure out the equation. I know u(x)=mgh and I tried to approximate for h, I'm not to sure about my answer but here is my work redone.

    http://img22.imageshack.us/img22/4167/cci2709200900000.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  7. Sep 28, 2009 #6
    yeah my work above was wrong, I was just overcomplicating things. U=-mgsin[tex]\theta[/tex]

    here is my work redone

    http://img297.imageshack.us/img297/6828/cci2809200900000.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  8. Sep 28, 2009 #7

    Redbelly98

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    Hi,

    You've made some more progress but there is still a problem with the U(x) expression.

    You had come up with
    U = mgh​
    which is correct.

    In your latest post, you also wrote

    U = mg sinθ​
    implying that h=sinθ, which is not correct. Also, the units do not work out: U has units of energy, while mg·sinθ has units of force. The two sides of any physics equation must have consistent units; if they do not then something is definitely wrong.

    So ... what is h, in terms of x and θ? That's what you need to work out, using some trigonometry. Then substitute that expression for h into
    U = mgh​
    Hopefully then it will work out.
     
  9. Sep 28, 2009 #8
    okay i got h=Lsin[tex]\theta[/tex] (I just replaced the x with L giving the wording of the question. So then U=mgLsin[tex]\theta[/tex] and I redid the work.

    http://img198.imageshack.us/img198/6828/cci2809200900000.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
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