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Spring constant problem

  1. Oct 25, 2004 #1
    A 0.515 kg wood block is firmly attached to a very light horizontal spring (k = 180 N/m) as shown in Fig. 6-40. It is noted that the block-spring system, when compressed 5.0 cm and released, stretches out 2.3 cm beyond the equilibrium position before stopping and turning back. What is the coefficient of kinetic friction between the block and the table?

    I am a little lost on this everyone.

    i thought i would use this formula and then solve for the Fr

    .5mv^2 + mgx +.5kx^2 = .5mv^2 + mgx +.5kx^2 +Frd

    once i have Fr then solve for the coefficient

    i cant seem to figure out what is wrong

    but i know the answer is .481
  2. jcsd
  3. Oct 25, 2004 #2


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    Dearly Missed

    ".5mv^2 + mgx +.5kx^2 = .5mv^2 + mgx +.5kx^2 +Frd"
    This is not meaningful; what is it supposed to mean????????
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