A spring of stiffness k supports a box of mass M in which is placed a block of mass m. If the system is pulled down a distancde d from the equilibrium position and released, find the force of reaction between the block and the bottom of the box as a function of time. Neglect any air resistance. For what value of d does the block just begin to leave the bottom of the box?(adsbygoogle = window.adsbygoogle || []).push({});

What i considered for this is as follows:

using newtons second law:

for the overall system,

[tex](m+M)a = kd - (m+M)g[/tex]

then for the block of mass m:

[tex]ma=F_{normal}-mg[/tex]

then i subbed in the second equation for ma in the first, and tried to rearrange for [tex]F_{normal}[/tex]

however i am not sure how to get the force as a function of time, writing a as dv/dt doesnt help since i cannot rearrange it to integrate i think.

thanks for the help

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# Homework Help: Spring and masses

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