# Mass within a box hung from a spring

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1. Oct 7, 2016

### Elvis 123456789

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. Oct 7, 2016

### BvU

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.

3. Oct 7, 2016

### haruspex

Are you defining up or down as positive?