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Spring constant help

  1. Oct 30, 2005 #1

    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
  2. jcsd
  3. Oct 30, 2005 #2


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    Show us some of your work!
  4. Oct 30, 2005 #3

    Ff= mu*n

    sum of Fx = -Ff + F
    sum of Fy = N+-w
    N= w
    therefore Ff = mu*m*g

    sum Fx = mu*m*g+F

    work done on a spring = (1/2)kx2^2-(1/2)kx1^2

    i'm stuck after i plug my equations in.
  5. Oct 30, 2005 #4


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    You know that the total amount of work done (by friction) is [itex]2 \mu mg \Delta x = \frac{1}{2} m v^2[/itex] where [itex]\Delta x[/itex] is the amount of compression the spring undergoes. When the spring is fully compressed, the total mechanical energy is just the potential energy of the spring (the object is at rest!) and all of that gets lost upon returning the starting point so that [itex]\frac{1}{2}k \Delta x^2 = \mu mg \Delta x[/itex].

    That should be enough information for you to obtain the value of the spring constant.
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