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Conservation of Energy Question

  1. Feb 1, 2017 #1
    1. The problem statement, all variables and given/known data
    Theres a pic.
    2. Relevant equations
    ##U_{spring}=U_s=\frac 1 2kx^2##

    3. The attempt at a solution
    I found (a) using ##W_{fr}=Δ(ME)## which ##v=7,4 \frac m s##
    For (b) I wrote;

    ##\frac 1 2kx^2-\frac 1 2mv^2=W_{fr}##
    ##\frac 1 2kx^2-\frac 1 2mv^2=-F_{fr}x## (here x are the same and represents the comprassion distance)
    so we get quadratic equation.
    I found ##\frac 1 2mv^2=48988J##
    so ##75000x^2+4400x-48988=0##
    I solved it but didnt came the answer where did I go wrong ?

  2. jcsd
  3. Feb 1, 2017 #2
    What did you get for an answer? And what is the right answer?
  4. Feb 1, 2017 #3
    I found 77 cm answer says 90 cm
  5. Feb 1, 2017 #4
    Sorry, it took me so long to get back. I was off doing something else. I got 0.78 m (or 78 cm) for my answer.
  6. Feb 1, 2017 #5
    then we get same results so there must be some roundings during the process...so I am correct I guess thank you
  7. Feb 1, 2017 #6
    I solved (c) but (d) is hard.Can someone help me ?
  8. Feb 1, 2017 #7


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    Did hitting the spring turn off gravity?
  9. Feb 1, 2017 #8
    Yeah...you are right.I didnt see that
  10. Feb 1, 2017 #9
    Me either.
  11. Feb 2, 2017 #10
    I found 0.93m when I did it.

    I found also (c) correctly with an exact result.

    Still having problem with (d)
  12. Feb 2, 2017 #11


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    Hmm... I think I got .90m.
    d) seems a bit strange to me. Haven't you already worked out all the distances moved in b) and c)? Should just be a matter of adding them up. It feels like it is hinting at a shortcut to get this answer without using the intermediate ones, but I don't think there is one.
  13. Feb 2, 2017 #12
    0.93 is approxemetly 0.90 I dont mind so much.

    The answer is 15m for (d).

    (b) is 0.9m and (c) is 2.8m
  14. Feb 2, 2017 #13


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    Consider the initial position and the eventual final (resting) position of the cab. The total energy input and eventually dissipated is determined by these points...
  15. Feb 2, 2017 #14
    I guess I see...Last energy of our system is 47628J And he took 2.8m+0.9m=3.7m till now.He will go up and down until loses all energy which is 47628J.
  16. Feb 2, 2017 #15


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    How silly of me.
  17. Feb 2, 2017 #16
    Then ;

    here D comes 14.8m ?? I it this the answer
  18. Feb 2, 2017 #17
    What are you using for the value of d? Are you including the displacement of the elevator to reach the equilibrium point?
  19. Feb 2, 2017 #18
    d is in the given part.3.7 m.The block will stop there
  20. Feb 2, 2017 #19
    Will it?
    Disclaimer: I have been wrong so many times that I question my own judgment. :)
  21. Feb 2, 2017 #20
    I think that way cause.That seems logical to me..lets wait for another opinion
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