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A Mass-Spring System with Recoil and Friction

  1. Dec 10, 2004 #1
    Here's the question:
    "An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction mu between the object and the surface. The object has speed v when it reaches x = 0 and encounters a spring. The object compresses the spring, stops, and then recoils and travels in the opposite direction. When the object reaches x = 0 on its return trip, it stops."

    "Find k , the spring constant. Express in terms of mu, m, g, and v."

    I'm stuck. So I used the conservation of energy law + the friction work. They don't want x in the terms of the answer. How can that be possible when x is part of the elastic potential energy.

    Any helps would be appreciated. :confused:
  2. jcsd
  3. Dec 10, 2004 #2
    When the object touches the spring you have :
    [tex]{\mu}N + kx = ma_x[/tex] and [tex]N -mg = 0[/tex]

    So the normal force is always N =mg because you don't move in the y-direction.

    When the object is recoiled you have in the x-direction :

    [tex]{\mu}N - kx = ma_x[/tex]

    Try to use these equations...

    You can't use conservation of energy because due to the friction, this system is NOT conservative...

    marlon, ps : N is the normal force on the object in the y-direction...
  4. Dec 10, 2004 #3
    Thanks, Marlon

    but wouldn't I still be stuck with x, if I use the (mu)mg - kx = ma equation?

    Also, do I use 0 = v^2 + 2ax to get rid of a? (vf = 0) :confused:
    and have a = -(v^2)/2x, which I 'm stuck with x? but they want the answer in terms of mu, m, g, and v.
    Last edited: Dec 10, 2004
  5. Dec 10, 2004 #4
    I suggest you work with [tex]x = x_0 + v_0 * t + \frac{at^2}{2}[/tex]

    Also make use of the conditions that are given... v at x = 0 etc...

  6. Dec 10, 2004 #5
    but if I use that equation, wouldn't there be one more variable (t) involved that's not part of the terms they want; wouldn't that make things even more complex?
  7. Dec 10, 2004 #6
    No, i just gave a push into the right direction...it is not the whole picture...

    Look how about this : x = vt and v = at...if you substitute t then this yields : v²= ax

  8. Dec 10, 2004 #7
    are you saying use a=(v^2)/x

    and (mu)mg - kx = m(v^2)/x ? This equation still has x in it and they only want the answer in terms of mu, m, g, and v.

    and if you say x = (v^2)/a, then it's like going around in circle with either a or x as part of the terms for the answer, but they neither want x, nor a.

    I'm confused. :confused: How could it be possible to find k in terms of only mu, m, g, and v?
  9. Dec 10, 2004 #8
    (mu)mg - kx = m(v^2)/x

    If I rearrange the above equation, I get kx = (mu)mg - m(v^2)/x
    Multiply the equation by x, I get k = mumgx - m(v^2)
    if x = 0, k = -mv^2

    That can't be right :yuck:

    Any helps would be appreciated. I'm really desperate now.
  10. Dec 10, 2004 #9
    Also, you can work with [tex]{\Delta}K + {\Delta}U +{\mu} * d = 0[/tex]

    Where K is kinetic energy...
    U potential energu of spring
    d displacement during the period when work is done...

    Evaluate these quantities ab initio and at the end...I mean one time when the object touches the spring + when the spring is maximally pressed down by the object. Then another time when the spring starts to move again and when the object needs to be still and the spring has again its natural length...

    good luck

  11. Dec 10, 2004 #10
    I've been stuck on the same problem for hours. :mad:
    They ask for the answer in terms of mu, m, g, and v only which is what really screws me up.
    I know there are all ready a lot of posts but any other help would be great.
  12. Dec 10, 2004 #11
    Finding the compression of the spring, I found that E_final=E_initial+W_nonconservative where E_final is zero and E_initial=(1/2*m*v^2).
    What would the W_nonconservative be?
  13. Dec 10, 2004 #12
    Any other help would be greatly appreciated :smile:
  14. Dec 11, 2004 #13
    How about writing a general equation for the object
    ma-kx = mu N where a = d^2 x/dt^2. Solving this DE will give t as a function of x. Substitute that in x = x0 + ut + 1/2 a t^2 to get the distance travelled x.
  15. Dec 11, 2004 #14
    Assuming I'm right, I figured k= (2*m*g*mu)/x but I need to get rid of x somehow so that the answer is in terms of mu, m, g, and v only...does anyone have any suggestions? :redface:
  16. Dec 11, 2004 #15
    I found the spring constant to be k=m*(((2*(mu)*g)/v)^2) but I'm not sure if my algebra is correct. Can anybody help? :confused:
  17. Dec 12, 2004 #16

    Doc Al

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    Staff: Mentor

    As marlon suggested, use energy methods to solve this. Assuming that the spring compresses to the point x = d. You can write two energy relationships:
    one for the compression to x = d:
    [tex]1/2 m v^2 = \mu mgd + 1/2 k d^2[/tex]
    another for the decompression back to x = 0:
    [tex]1/2 k d^2 = \mu mgd[/tex]
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