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Spring & Oscillation

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating on a large spring. Suppose an astronaut attaches one end of a large spring to her belt and the other end to a hook on the wall of the space capsule. A fellow astronaut then pulls her away from the wall and releases her. The spring's length as a function of time is shown in the figure

    http://session.masteringphysics.com/problemAsset/1073872/3/knight_Figure_14_36.jpg
    What is her mass if the spring constant is 240N/m
    2. Relevant equations

    Hookes Law
    F = -kx
    U = (1/2)*k*x2

    3. The attempt at a solution

    So the spring was pulled back 1.4 metres and then released where it went to the point 0.6
    so 0.6 is the point at equilibrium, so -0.8 is the displacement and that is the x

    F = (240 * 0.8)
    and then im not sure what to do, im not even sure if im doing this right and help would be appreciated
     
  2. jcsd
  3. Mar 29, 2009 #2
    im pretty sure my approach is wrong...

    F = -260 * -0.8 = 208N

    mg = weight

    208/9.8 = m

    m = 21.224489795918367346938775510204kg

    im sure thats wrong is there a better way to approach this problem and why is the way im approaching it wrong?
     
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