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Spring problem

  1. Oct 1, 2004 #1
    a block of mass M is held on a horizontal table against a compressed horizontal (massless) spring. When released from rest, the block is launched along the tabletop and eventually comes to a stop after sliding a total distance L. The initial spring compression is x_0 and the spring constant is K. Calculate teh coefficient of kinetic friction between the block and the ttable top in terms of the variables M, L, K, and X_0.

    Im lost. I know F=ma. I know x_0*k is the force the spring is dishing out. I know that the frictional force is Mu*M*g. I tried to put it together with v^2 = v_0^2 + 2aD. The problem with that is I dont know the initial velocity.

    How do I use X_0*K to calculate velocity?
  2. jcsd
  3. Oct 1, 2004 #2


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    Use an equation relating the mechanical energy of the system at 2 different places and the work of friction done over that interval.
  4. Oct 1, 2004 #3
    Could you walk me through that . . . I'm still lost.
  5. Oct 1, 2004 #4


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    OK, let's start with Newton's 2.law, and derive the energy equation.
    1.Newton's 2.law
    Here, "x" is the compressed length of the spring, and
    ([tex]t_{L}[/tex] is the time when the system stops; when a distance L has been traversed)
    [tex]-\mu{Mg}[/tex] is the frictional force, whereas a is the acceleration of the system.

    2. Derivation of energy equation
    We a) multiply the above equation with velocity v, and
    b) integrate from t=0 to [tex]t=t_{L}[/tex]:
    a) [tex]-Kxv-\mu{Mg}v=Mav[/tex]
    b) [tex]-\frac{K}{2}x(t=t_{L})^{2}+\frac{K}{2}x(t=0)^{2}-\mu{MgL}=\frac{M}{2}(v(t=t_{L})^{2}-v(t=0)^{2})[/tex]
    Or, by recognizing:
    we gain by rearranging:
    Last edited: Oct 1, 2004
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