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Work and Energy with Tension

  1. Jul 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Refer to figure.

    2. Relevant equations
    U=Fs
    T=.5mv^2
    Vg=mgh

    3. The attempt at a solution
    I am not exactly sure where to start. I would appreciate if someone could nudge me in the right direction and then go from there.
     

    Attached Files:

  2. jcsd
  3. Jul 15, 2016 #2

    haruspex

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    Start with a free body diagram of the mass when theta is 180. Write out the ΣF=ma equation for it.
     
  4. Jul 15, 2016 #3
    I have done that but I am not sure how to get the velocity of the mass...
     

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  5. Jul 15, 2016 #4

    haruspex

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    Your diagram seems to be for theta = 90 degrees.

    For the velocity, any conservation law come to mind?
    (In your 'relevant equations', I don't understand Vg=mgh. What is Vg there?)
     
  6. Jul 15, 2016 #5
    yeah my figure is incorrect. Vg is the gravitational potential energy
     
  7. Jul 15, 2016 #6

    haruspex

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    Ok. I thought it more usual to use E or U for energy. What is the T in the preceding equation? Well, I can guess from the equation what it is, but why 'T'?
     
  8. Jul 15, 2016 #7
    kinetic energy. it's just the notation we use.
     
  9. Jul 15, 2016 #8

    haruspex

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    Ok, so what relates that to Vg?
     
  10. Jul 15, 2016 #9
    U=ΔT-ΔV where V=Vg+Ve and Ve is the elastic potential energy(spring) but it isn't applicable to this problem
     
  11. Jul 15, 2016 #10

    haruspex

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    Let's investigate that. No spring here, so Ve is zero. What is U? Can you assign a value to it?
     
  12. Jul 15, 2016 #11
    U=ΔE=(T2-T1)+(VG2-VG1)
     
  13. Jul 15, 2016 #12

    billy_joule

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    That's right.
    What is the value of ΔE? In other words, does any energy enter or exit the system? (you are expected to assume the nudge does not add any kinetic energy to the pendulum).
     
  14. Jul 17, 2016 #13
    There are no external forces so now work is done on the system causing E=0. I figured out how to work in on my own. Thank you for your reply.
     
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