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Cambridge Physics Problems 10

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    This is a part of the "Cambridge Physics Problems" thread, an exercise which I have taken up myself. Please read here for a short introduction on the types of questions featured in threads of the same name. Please search for "Cambridge Physics Problems" for threads featuring past questions.
    I appreciate all of your help on this. If possible, please show me the mathematical workings of the question (since this is what the question emphasises).
    Thank you!

    A mass of 1.0kg is firmly attached to a rigid beam by a copper wire and is supported so that the wire is vertical but unstretched. The support is removed. Show that, if Hooke's law is obeyed, only half of the potential energy lost by the mass remains as elastic potential energy stored in the wire.

    The mass is now raised to the level of the rigid beam and released. Find the minimum cross-sectional area of the wire if it is not to break. Assume that the extension is always much less than the unstretched length of the wire and that Hooke's law is always obeyed.

    For copper the Young modulus, E = 1.1E11 Pa, tensile strength (maximum tensile stress), \tau = 3.0 x 10^8Pa.

    For the first part, is there any other way to show the required equation without calculus? I've tried the methods in Cambridge Physics Problems 9 but I failed. I will post my working for the second part soon.

    Thank you.
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    hey john,

    for the first part: Click image for larger version. 

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    If we just let the wire go it will oscillate with S.H.M so I'm assuming the wire is dampened. Hence it will eventually settle at its equilibrium position. At this equilibrium position 1g = T = kx (Eq 1.)

    Hence in terms of energy, 1gx is the Egp lost dropping to Equilibrium but 0.5kx^2 is the Eep gained

    Now looking back at Eq 1 we can substitute back for g to give the Gpe lost in terms of k and x as kx^2 which is twice evenrgy stored as Eep at equilibrium.

    EDIT: hope this helps! no calculus required. Also oops for forgetting the tension in my diagram
    Attached Thumbnails
    Click image for larger version. 

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    (Original post by Lord of Ruin)
    Now looking back at Eq 1 we can substitute back for g to give the Gpe lost in terms of k and x as kx^2 which is twice evenrgy stored as Eep at equilibrium.
    Am I thinking too much or is the question really that easy? The book (published in the late 90s) did mention the questions are for students preparing for STEP Physics and undergraduate degree. I feel so stupid right now.
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    As for the second part, I think it's again deceptively simple:

    Click image for larger version. 

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    I'm assuming when they say the height of the rod they mean the natural length correct?

    Young's modulus never came into my calculations. I find that strange and a source of possible worry about the validity of my working.

    Hope this helps!
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    (Original post by Lord of Ruin)
    As for the second part, I think it's again deceptively simple:

    Click image for larger version. 

Name:	copperwire2.jpg 
Views:	26 
Size:	236.0 KB 
ID:	143042

    I'm assuming when they say the height of the rod they mean the natural length correct?

    Young's modulus never came into my calculations. I find that strange and a source of possible worry about the validity of my working.

    Hope this helps!
    I'm sorry but your method isn't correct as well. The answer for the minimum cross-sectional area to prevent breaking is 2.4E-5 m^2.

    Can anyone else help us out with this? Thanks in advance.

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