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Force Constant, Springs

  1. Feb 26, 2008 #1
    Hi, I Need to know exactly how to solve this...I just can't take it anymore. The only reason I'm so into it now is because I don't get it. I promised my girlfriend I would help her and now shes annoyed because I get more into it than she does. But anyway, someone..PLEASE tell me what has to be done here. I'm sure its something simple but I have no more brain power to even remotely try.



    Two blocks, each with a mass 0.40 kg, can slide without friction on a horizontal surface. Initially, block 1 is in motion with a speed v = 1.2 m/s; block 2 is at rest. When block 1 collides with block 2, a spring bumper on block 1 is compressed. Maximum compression of the spring occurs when the two blocks move with the same speed, v/2 = 0.60 m/s.

    If the maximum compression of the spring is 1.9 cm, what is its force constant?




    I came up with 111N/m but only because I followed a patern, didn't feel like I knew what I was doing...is that right? 111?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 26, 2008 #2
    Well, how did you get that answer; helping you will be easier if we know your way of solving and possible errors in it.
     
  4. Feb 26, 2008 #3
    It would have been better if you had posted your work.

    Need some water ??


    anyways, its Energy conservation.

    [tex]\frac{mv^2_1}{2}=2 \frac{mv^2_{common}}{2}+\frac{Kx^2}{2}[/tex]

    K is force constant, x is compression.
     
  5. Feb 26, 2008 #4
    I set the change of potential energy equal to the change in kinetic energy. Then I took final velocity .6m/s and subtracted initial velocity 1.2m/s and set the product of that to equal force constant times the amount the spring collapsed. Having all those values, I calculated for "k".

    ...?
     
  6. Feb 26, 2008 #5
    I'm sorry, but I didn't get you properly. Can you post your work here.
     
  7. Feb 26, 2008 #6
    {mv^2_{common} - Would this be the value 2/v = .60m/s ? and in terms of mass, do I just double the mass of each box?

    Oh, and Sourbah....putting up my work would not help, it's most definatley wrong. I just basically guessed...not much learning there.
     
  8. Feb 26, 2008 #7
    Use the equation which google spider putted up in post #3, I'm not sure you are using the same equation. If you still don't get, just post again with your difficulty
     
  9. Feb 26, 2008 #8
    I have trouble rearranging the variables to solve for the one I need. Also, I don't understand what mv^2common is.....

    What will this equation look like with my numbers plugged in. I want to work at backwards, understand it conceptually....
     
  10. Feb 26, 2008 #9
    [tex] \frac{mv^2_{common}}{2}[/tex] is the mass of each body (common in this case) multiplied by the velocity (common at the instant of maximum compression); its the KE at the time of maximum compression.

    Just write up the values of those variables in the equation which you know, we'll see what remains and how can it be calculated (You may feel irritated with my replies but remember, once I tell you the solution, the fun will be over)
     
  11. Feb 26, 2008 #10
    I'm not irritated, I've already gone through that stage...I'm just in my "got to have the answer" so I can figure out what had to be done to get it. I don't know what the velocity was at the moment of maximum compression. I don't know how to calculate to get those values.

    I have:

    (.4kg)(1.2m/s)^2 / 2 = 2(.8kg)(1.2) /2 + K(.019m)/ 2

    How does that look to you?
     
  12. Feb 26, 2008 #11

    tiny-tim

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  13. Feb 26, 2008 #12
    thank you tiny tim....i remember. Really, if there's something that's troubling me...it's very hard for me to get out of the cycle, like my brain shuts down. If I had the answer, and the other values which I don't know...i could work backwards and make perfect sense of it all...
     
  14. Feb 26, 2008 #13
    Dude, that should look like this :
    (.4kg)(1.2m/s)^2 / 2 = 2(.4kg)(0.6)^2 /2 + K(.019m)^2 / 2
     
  15. Feb 26, 2008 #14
    Thanks to everyone who offered their help today... I've spent a week reading a physics book and have been very into it. I'm just doing it for myself so that I could help my friends and family with it if need be. It's a lot harder than I expected but hopefully with some help myself I'll be able to get on top of it soon.
     
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