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?