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Extension in a spring

  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data

    Scenario: a block of mass 5 kg is hanging from a spring from an ideal pulley from one side, the other side supporting a mass of 10 kg through a STRING.
    Now, the 10 kg block replaced with a 5 kg block.

    In which case would the extension in spring be greater, assuming constant acceleration.
    2. Relevant equations
    10(g-a)=5(g+a) - kx1
    and
    5(g+a)-kx2= 5(g-a)
    3. The attempt at a solution
    I know the answer is the first case, visualising it(and according to the solution). But i'm not able to prove this mathematically. From my equations i'm getting x1 proportional to (15a-5g) and x2 proportional to 10a and i don't know how these are related.
    Thanks for the help :-)
     
    Last edited: Feb 26, 2017
  2. jcsd
  3. Feb 26, 2017 #2

    kuruman

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    What spring? Your scenario says nothing about a spring.
     
  4. Feb 26, 2017 #3
    my bad i'm so sorry. the spring is connected to the block.
     
  5. Feb 26, 2017 #4

    CWatters

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    That description doesn't make any sense to me. Any chance of a diagram?
     
  6. Feb 26, 2017 #5
    s4.gif
    this is the second case.
     
  7. Feb 26, 2017 #6

    kuruman

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    Can you find the acceleration in each of the two cases?
     
  8. Feb 26, 2017 #7
    how can i? i mean, it's related to k and extension in the spring. sorry if i'm missing something very common.
     
  9. Feb 26, 2017 #8

    kuruman

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    Let's look at the second case first with the two 5-kg masses. What do you think the acceleration is in this case?
     
  10. Feb 26, 2017 #9
    i use the equation 5(g+a)-kx = 5(g-a) to get a = kx/10.
     
  11. Feb 26, 2017 #10

    kuruman

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    Wherever you got that equation, it does not apply here. Suppose I hid the spring with a piece of cardboard (see figure) so that you are not able to see it. All you see is two equal masses on either side of the pulley. You don't know the spring is there and you assume that the string is continuous behind the cardboard. Which way do you think the pulley will be accelerating under these circumstances? Clockwise or anticlockwise?
    Ameya_3.png
     
  12. Feb 26, 2017 #11
    wouldn't a be 0 if we assume no spring?
     
  13. Feb 26, 2017 #12

    kuruman

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    Exactly. It is zero. What is the tension in the left string? What about the right string?
     
  14. Feb 26, 2017 #13
    5g in both.
     
  15. Feb 26, 2017 #14

    kuruman

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    Very good. Now suppose I remove the cardboard and you discover that there is an extended spring behind it. Would anything change? Would the masses all of a sudden accelerate?
     
  16. Feb 26, 2017 #15
    okay, so the extension will be kx=mg so x =mg/k. am i right?
    and was the equation for my first case right?
     
  17. Feb 26, 2017 #16

    kuruman

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    That is correct.
    One thing at a time. I hope you have seen by now that the tension in the string is the force that causes the extension. So now you need to find the tension in the first case.
     
  18. Feb 26, 2017 #17

    haruspex

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    A massless spring exerts the same force at each end. You seem to have taken it as only exerting a force at one end.
     
  19. Feb 27, 2017 #18
    l don't understand, could you please explain further?
     
  20. Feb 27, 2017 #19
    I get T = 10(g-a) = kx. is that right? but this has 'a' in it's equation, so how would i compare it to the second case?
     
  21. Feb 27, 2017 #20

    haruspex

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    Ok, but it will be easier if you first explain how you get
     
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