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Block can slide along a table where the coefficient of friction

  1. Dec 12, 2004 #1
    A 198g wood block is firmly attached to a very light horizontal spring, as shown in the figure below.





    The block can slide along a table where the coefficient of friction is 0.287. A force of 23.1 N compresses the spring 17.7 cm. If the spring is released from this position, how far beyond its equilibrium position will it stretch on its first swing

    the way i approached this:

    1/2kx^2 - Ukmgx = 1/2mv^2 = UkmgD
    where D is the distance it travels after acquiring kinetic energy.

    however, there is something wrong with this approach and i cannot figure out what it is. please help!
     
  2. jcsd
  3. Dec 12, 2004 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    You are omitting the spring potential energy when the mass has extended to the distance D past the equilibrium position. This is provided by the original spring potential which also provides the work done against friction.

    [tex]\frac{1}{2}k(x^2-D^2) =\mu_kmg(x+D)[/tex]

    AM
     
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