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Spring and box problem

  1. May 6, 2010 #1
    A 8 kg box slides d = 4.4 m down the frictionless ramp shown in Figure CP10.69, then collides with a spring whose spring constant is 250 N/m. At what compression of the spring does the box have its maximum velocity?


    i know that .5kx^2+.5mv^2=mg(d+x)sin theta
    -------------spring--kinetic-----potential----

    but what is next?
     

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  2. jcsd
  3. May 6, 2010 #2

    tiny-tim

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    Hi xstetsonx! :smile:

    (have a theta: θ and try using the X2 tag just above the Reply box :wink:)
    That's fine … now put v on the LHS, and maximise the RHS. :smile:
     
  4. May 6, 2010 #3
    i am sorry LHS (left hand side)? if so v is already on the left hand side and how i max the right?
     
  5. May 6, 2010 #4

    tiny-tim

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    I meant on its own.
     
  6. May 6, 2010 #5
    but how do i know what x is?
     
  7. May 6, 2010 #6

    tiny-tim

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    uhh? :confused: x is distance … you have a function of x, and you need to maximise that function
     
  8. May 6, 2010 #7
    how do i do that?
    do i take the derivative????
     
    Last edited: May 6, 2010
  9. May 6, 2010 #8

    tiny-tim

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    That's right … it's a maximum (or minimum or inflection point) when the derivative is zero. :smile:
     
  10. May 6, 2010 #9
    OMG that IS why my solution manual did the derivative....... this is back to calculus isn't it don't you just hate they tell you the solution without saying why


    o btw do i do it in respect of x or v?
     
  11. May 6, 2010 #10

    tiny-tim

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    well, duh … it's a spring question …

    so they spring it on you! :biggrin:
    v is a function of x, and you want to maximise v,

    so you differentiate v wrt x :smile:

    (if you draw a graph, with v going up and x going across, then dv/dx is the slope, so the maximum for v will be when the slope is horizontal, ie zero)
     
  12. May 6, 2010 #11
    god this problem sucks i still haven't been able to solve my equation and get the right answer-----(15.68)
     
  13. May 6, 2010 #12

    tiny-tim

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    hmm … i make it .01568 :confused:

    what equation did you get?
     
  14. May 6, 2010 #13
    oops it is in cm and yours i think is in m. so ur answer is right. but do u mind show me how you get there?
     
  15. May 6, 2010 #14

    tiny-tim

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    erm :redface:you show us how you got your answer
     
  16. May 6, 2010 #15
    i didn't get it it comes from the solution manual .....
     
  17. May 6, 2010 #16

    tiny-tim

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    Yes, I know that, but you said you tried and got the wrong answer …
    … so show us what you did. :smile:
     
  18. May 6, 2010 #17
    v=((-125x^2+39.2x+172.48)/(4))^.5

    dv/dx=.5(-31.25x^2+9.8x+43.12)^-.5(62.5x+9.8)

    definitely don't think that is right
     
    Last edited: May 6, 2010
  19. May 7, 2010 #18

    tiny-tim

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    Hi xstetsonx! :smile:

    (just got up :zzz: …)
    (please use the X2 tag just above the Reply box :wink:)

    what's wrong with that? (apart from the missing minus sign)

    you now have dv/dx = 0 when x = 9.8/62.5 :smile:

    (also, maximising v is the same as maximising v2, so you could simply have gone for dv2/dx = 0 instead :biggrin:)

    btw I should have mentioned before :redface: … if you don't like differentiating, you can just complete the square instead :wink:
     
  20. May 7, 2010 #19
    sorry didn't have time to reply till now

    the x^2 hates me it didn't work..... or i don't know how to use it

    (please use the X2 tag just above the Reply box :wink:)



    do you set that to 0???? and solve for x? and then.....wat?



    what's wrong with that? (apart from the missing minus sign)

    you now have dv/dx = 0 when x = 9.8/62.5 :smile:

    (also, maximising v is the same as maximising v2, so you could simply have gone for dv2/dx = 0 instead :biggrin:)



    complete the square in what step?




    btw I should have mentioned before :redface: … if you don't like differentiating, you can just complete the square instead :wink:[/QUOTE]




    guess i don't know how to quote either......
     
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