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I would like to get a confirmation about this one

  1. Sep 1, 2006 #1
    Hello!

    I think I solved the following problem but the answer (given by my professor) does not match with mine. Therefore, I would be happy if someone could confirm me that he is wrong ;)

    There is a spring attached to a box of 200 gr. mass. The box is lifted up so that the lower end of the spring is 30 cm above the ground. Suddenly, the box is falling down to the ground. What is the spring's compression through the impact (s) ?
    The spring's constant (k) is 156,96 N/m.

    I went this way:

    Duration of the fall:
    h= 0.3 m
    [tex] t= \sprt{\frac{2*h}{g}} = 0.247 s [/tex]

    with the duration I get the final speed of the box:

    [tex] v= gt = 2,43 \frac{m}{s} [/tex]

    My thougt it that the potential energy of the spring has to be the amount of the kinetec energy and the potential energy before the impact:

    [tex] \frac{1}{2}mV^2+mgh = \frac{1}{2}ks^s [/tex]

    this equation gives me the compression (s) :

    [tex] \sqrt{\frac{mV^2+2mgh}{k}}= 0.1225 m [/tex]

    The answer given by my professor is 0,1 m.

    Do you think he made a mistake as well ?

    Thanks ! fara
     
  2. jcsd
  3. Sep 1, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Nothing wrong with this. This is the speed of the box at the instant the spring just touches ground.

    Good!
    How does "h" relate to "s"? (The h here is not .3 m!)

    You are mixing up this "h" with the previous height of .3 m, I bet.

    An easier method is use conservation of energy right from the start. Hint: What is the change in height of the box as it drops to its lowest point?

    The professor's answer is correct, by the way. :wink:
     
    Last edited: Sep 1, 2006
  4. Sep 1, 2006 #3

    Andrew Mason

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    Science Advisor
    Homework Helper

    You have to assume a massless spring.

    Find the equilibrium point of the spring (where compression force = gravity: mg = kx). Any compression beyond that point is due to the energy from the fall.

    So:

    [tex]mg = kx_{eq}[/tex]

    [tex] E = mg(h - (l - x_{eq}))[/tex] where l is the length of the spring and h is the height of the box (h-l = 30 cm).

    equate that to the spring energy at maximum compression (ie compression beyond [itex]x_{eq}[/tex]):

    [tex] E = \frac{1}{2}kx_{max}^2[/tex]

    and solve for [itex]x_{max}[/itex]



    AM
     
    Last edited: Sep 1, 2006
  5. Sep 1, 2006 #4
    Thank you very much guys!
    It is so logical if you think about it. The box does not go all the way to the ground, it will remain a certain distance from the ground. So it is important for a equilibrium that both energies are exactly the same. I learned a lot with this one!

    [tex] mg(h-s)=\frac{1}{2}ks^2 [/tex] that leads me to
    [tex] 0 = s^2 + \frac{2mgs}{k}-\frac{2mgh}{k} \rightarrow \\
    s= - \frac{mg}{k} + \sqrt{\frac{mg}{k}^2+\frac{mgh}{k}}= 0.0125 m + 0.0875 m = 0.1 m
    [/tex]
     
    Last edited: Sep 1, 2006
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