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Mass within a box hung from a spring

  1. Oct 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A block of mass m is placed inside a box of mass M , which is then hung from a spring with spring constant k. The system is pulled down some distance d and released at time t =0. Determine the reaction force between the block and the bottom of the box as a function of time. For what value of d will the block just lose contact with the bottom of the box when at the top of the vertical oscillation?

    2. Relevant equations
    F = ma
    y = Acos(ωt + φ)
    ω2 = k/m

    3. The attempt at a solution
    A free-body diagram will show for mass "m" leads to
    FR - mg = ma -----> FR = m(g + a)

    the acceleration for the block "m" and the box "M" will be the same

    so y = Acos(ωt + φ) ----> y = dcos(ωt) where ω = sqrt (k/(M + m) )

    differentiating twice gives

    a = -dω2cos(ωt)

    plugging back into the equation for the reaction force gives

    FR = m(g -dω2cos(ωt))

    for the second part set FR = 0 which gives

    d = g/[ω2cos(ωt)]

    does this look okay to you guys?
     
  2. jcsd
  3. Oct 7, 2016 #2

    BvU

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    No. PF isn't really in the business of stamp-approving homework.

    But at t=0 I get something less than mg for FR, which seems strange to me.
     
  4. Oct 7, 2016 #3

    haruspex

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    Are you defining up or down as positive?
     
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