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2 Potential Energy questions

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    Some of you may have noticed that I've been posting quite frequently for help with questions involving Potential/Kinetic Energy. It is a topic I've been struggling with a little, but this should be the last time for a while that I'll be needing help with it.

    1)
    A smooth wire is bent into the shape of the graph of  y=x+2 \sin x for  0<x<\pi , the units being metres. Points A, B and C on the wire have coordinates  (0,0), (\pi,\pi), (2\pi,2\pi) . A bead of mass m kg is projected along the wire from A with speed u ms^{-1} so that it has enough energy to reach B but not C. Prove that u is between 8.66 and 11.10.

    If u=10, the bead comes to rest at a point D between B and C. Find the greatest speed of the bead between B and D,

    AnswerSo I assume that if the bead has enough energy to reach B but not C, then it's fair to model that as mg2\pi>\frac {1}{2}mv^2>mg\pi. Doing the algebra gives me 2\sqrt{g\pi}>v>\sqrt{2g\pi} , giving me 11.10>v>7.85. So my upper value is correct, but not the lower one, and I'm quite confused as to why.

    For the next part, I assumed that the greatest speed of the bead between B and D would be the speed of the bead as it just passes B (as the graph of y=x+2\sin x always as a positive or flat gradient). The answer is apparantly  7.2 ms^{-1}, and I don't have any idea how they got that.


    So now for question 2
    2 A block of mass M is placed on a rough horizontal table. A string attached to the block runs horizontally to the edge of the table, passes round a smooth peg, and supports a sphere of mass m attached to its other end. The motion of the block on the table is resisted by a frictional force of magnitude F, where F<mg. The system is initally at rest.
    a) Show that when the block and the sphere have each moved a distance h, their common speed is given by v^2=\frac{2(mg-F)h}{M+m}
    b) Show that the total energy lost by the sphere as it falls through the distance h is \frac{m(Mg+F)h}{M+m}

    AnswerNow I can manage part a: Fs-Rs = Energy gain , so mgh-Fh = \frac {1}{2} (M+m)v^2. This is easy enough to rearrange to make v^2 the subject and get the expression in part a).

    Part b is giving me problems. I assume the energy lost is equal to  Fh, which is equal to mgh - \frac{1}{2}mv^2, but when I substitute in the answer for v^2 from part a) I just get a mess that no matter how I try to tidy up never looks anything like the expresssion I should be ending up with.

    Help would be extremely appreciated.
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    hmmm, I get 7.85 as well
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    If you graph the curve, you'll see that in order to reach B, it must go over a hump earlier on, which is higher than B.

    Edit: If you don't want to sketch the curve, the fact that the gradient of the curve at B is negative, will also tell you that there is a hump prior to B.
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    (Original post by ghostwalker)
    If you graph the curve, you'll see that in order to reach B, it must go over a hump earlier on, which is higher than B.

    Edit: If you don't want to sketch the curve, the fact that the gradient of the curve at B is negative, will also tell you that there is a hump prior to B.
    Ahhhhhhh I differentiated and sketched x + sinx so missed that
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    Can anyone help with the second one?
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    (Original post by Julii92)
    Can anyone help with the second one?
    Your start to 2b is correct and leads to the correct answer

    Energy = mgh - \dfrac{m(mg-F)h}{M+m}

    add fractions

    Energy = \dfrac{(mg(M+m)-m^2g+mF)h}{M+m}


    Energy = \dfrac{(mMg+m^2g-m^2g+mF)h}{M+m}


    ... ...
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    (Original post by TenOfThem)
    Your start to 2b is correct and leads to the correct answer

    Energy = mgh - \dfrac{m(mg-F)h}{M+m}

    add fractions

    Energy = \dfrac{(mg(M+m)-m^2g+mF)h}{M+m}


    Energy = \dfrac{(mMg+m^2g-m^2g+mF)h}{M+m}


    ... ...
    Right! Thank you (And Ghostwalker)

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Updated: August 3, 2012
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