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Relationship between Velocity, Kinetic Energy and Mass

  1. Jun 3, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-6-3_3-53-33.png
    upload_2016-6-3_4-10-39.png

    2. Relevant equations
    i) F=ma=mv/t
    ii) E=mv2/2

    3. The attempt at a solution
    Now based on equation 1 , I have concluded that the velocity gained is inversely proportional to m, and that so is the kinetic energy , I have seen otherwise.
     
  2. jcsd
  3. Jun 3, 2016 #2

    BvU

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    How can you conclude that if equation 1 does not contain any distance/displacement at alll ? You need other equations !
     
  4. Jun 3, 2016 #3
    Oh hmmmm, x=at^2/2, which means, a=2x/t^2, which means that mv/t=2mx/t^2=F, so does this mean they are independent , I am sorry , I am a little weak.Could you please elaborate?
     
  5. Jun 3, 2016 #4

    BvU

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    mv/t = F doesn't help you if you don't know how t depends on m.

    Would the exercise be easier for you if the text would read:

    You have a force ##F## to accelerate a mass ##m## from rest with constant acceleration over a distance ##d##. What is the final velocity ?​
     
  6. Jun 3, 2016 #5
    I think I see something , the energy gained by the system would be Fd=mad, so this implies that mv^2/2=mad, v^2/2=ad, so this would mean the kinetic energy is independent of m ? Further a=F/m , v=sqrroot(Fd/m), does that mean v varies in an inverse proportion with sqrt(m)
     
  7. Jun 3, 2016 #6
    Sir ?
     
  8. Jun 3, 2016 #7

    BvU

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    Correct
    Bingo.

    Can you also do the alternative route ? ## \ d = {1\over 2} at^2\ ## tells you ##\ t \propto \sqrt d\ ## and you already had ##\ mv/t = F ##
     
  9. Jun 3, 2016 #8
    Thank you so much sir , I get it now
     
  10. Jun 3, 2016 #9
    Hmmmmm
     
  11. Jun 3, 2016 #10
    Right , I get back the same place sir, thank you so much sir , that helped alot.
     
  12. Jun 3, 2016 #11

    BvU

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    Hmmm as in "delicious" or Hmmm as in "I don't belive a word of what I read" :smile: ?
     
  13. Jun 3, 2016 #12
    lol "hmmm" as in intriguing
     
  14. Jun 3, 2016 #13

    BvU

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    Ah, posts crossed. Well done. On to the next exercise.

    PS
    You have a force ##F## to accelerate a mass ##m## from rest with constant acceleration over a distance ##d##. What is the final velocity ?​
    was meant to seduce you to calculate v as a function of knowns:
    $$\ d = {1\over 2} at^2\ \Rightarrow t = \sqrt{2d\over a} \Rightarrow v = at = {F\over m} \sqrt{2dm\over F} \Rightarrow v\propto 1/\sqrt m$$
     
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