Question

A block of mass M rests on a table. It is fastened to the lower end of a light, vertical spring. The upper end of the spring is fastened to a block of mass m. The upper block is pushed down by an additional force 3 mg, so the spring compression is $\mathbf{4}\mathbf{mg}/\mathbf{k}$. In this configuration, the upper block is released from rest. The spring lifts the lower block off the table. In terms of m, what is the greatest possible value for M?

Short answer

The greatest possible value for $\mathrm{M}=2\mathrm{m}$.

Step-by-step solution

Step 1:

Isolated system (energy), the total energy of an isolated system conserved,

${\mathbf{\Delta E}}_{\mathrm{mech}}\mathbf{=}\mathbf{0}$

This can be written as follows:

role="math" localid="1663597401666" $\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{+}{\mathbf{\Delta E}}_{\mathrm{int}}\mathbf{=}\mathbf{0}$

Here,

$\mathbf{\Delta U}\mathbf{=}$Change in potential energy

$\mathbf{\Delta K}\mathbf{=}$Change in kinetic energy

${\mathbf{\Delta E}}_{\mathrm{int}}\mathbf{=}$Change in internal energy

Step 2:

As the lower block is just lifted from the table, the upward force applied to it must be equal to the weight of the block. The extension of the spring, from, $\left|{\overline{\mathrm{F}}}_{\mathrm{s}}\right|=\mathrm{kx}$must be $\frac{\mathrm{Mg}}{\mathrm{k}}$. From point at release to the point when the moving block first comes to rest, apply the law of conservation of energy:

$\begin{array}{c}{\mathrm{K}}_{\mathrm{i}}+{\mathrm{U}}_{\mathrm{gi}}+{\mathrm{U}}_{\mathrm{si}}={\mathrm{K}}_{\mathrm{f}}+{\mathrm{U}}_{\mathrm{gf}}+{\mathrm{U}}_{\mathrm{sf}}\\ 0+\mathrm{mg}\left(-\frac{4\mathrm{mg}}{\mathrm{k}}\right)+\frac{1}{2}\mathrm{k}{\left(\frac{4\mathrm{mg}}{\mathrm{k}}\right)}^2=0+\mathrm{mg}\left(\frac{\mathrm{Mg}}{\mathrm{k}}\right)+\frac{1}{2}\mathrm{k}{\left(\frac{\mathrm{Mg}}{\mathrm{k}}\right)}^2\end{array}$

$\begin{array}{c}-\frac{4{\mathrm{m}}^2{\mathrm{g}}^2}{\mathrm{k}}+\frac{8{\mathrm{m}}^2{\mathrm{g}}^2}{\mathrm{k}}=\frac{{\mathrm{mMg}}^2}{\mathrm{k}}+\frac{{\mathrm{M}}^2{\mathrm{g}}^2}{2\mathrm{k}}\\ 4{\mathrm{m}}^2=\mathrm{mM}+\frac{{\mathrm{M}}^2}{2}\end{array}$

$\begin{array}{c}\mathrm{mM}+\frac{{\mathrm{M}}^2}{2}+4{\mathrm{m}}^2=0\\ \mathrm{M}=\frac{-\mathrm{m}\pm \sqrt{{\mathrm{m}}^2-4\left(\frac{1}{2}\right)(-4{\mathrm{m}}^2)}}{2.\frac{1}{2}}\\ \mathrm{M}=-\mathrm{m}\pm \sqrt{9{\mathrm{m}}^2}\end{array}$

Mass cannot be negative. So,

$\begin{array}{c}\mathrm{M}=\mathrm{m}(3-1)\\ =2\mathrm{m}\end{array}$.

The value of mass$\mathrm{M}$ is twice that of m.