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Displacement of spring of a block going down an incline

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data
    An inclined plane of angle θ = 20.0° has a spring of force constant k = 455 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.71 kg is placed on the plane at a distance d = 0.330 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?


    2. Relevant equations
    dE=dk+du=0

    Kf-Ki+Dfsp-Disp+Dfg-Dig=0
    Ki=1/2mv^2
    Dfsp=1/2kx^2
    Dig= mg[h + x(sin20.0)]


    3. The attempt at a solution
    Kf,Disp, and Dfg all Equal zero
    so by substitution:

    -1/2mv^2+1/2kx^2-mg[h + x(sin20.0)]=0
    -(2.71)(.75^2)/2+455/2x^2-(2.71*9.8)[0.33+xsin(20)]=0

    which simplifies to
    227.5x^2-9.084x-9.52=0

    My answer is .226, which is the incorrect answer. What am I doing wrong?

    Thanks!
     
    Last edited: Oct 12, 2011
  2. jcsd
  3. Oct 13, 2011 #2
    I forgot to post the picture. Here it is:
    7-p-063.gif
     
    Last edited: Oct 13, 2011
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