Question

A system consists of two cubes of mass m1, and m2 respectively connected by a spring of force constant k. force (F) that should be applied to the upper cube for which the lower one just lifts after the force is removed, is

Answer 1

Answer:

F=mg  

So ,here m=M1+M2

There for F=(M1+M2)g

Answer 2

Answer:

The force (F) that is applied to the upper cube for which the lower cube rises right after the force has been removed, is $\bold{ \mathrm{F}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{g} }$.

Explanation:

Initially in the state of equilibrium:

$\begin{aligned}&\mathrm{F}+\mathrm{m}_{1} \mathrm{~g}=\mathrm{kx}_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}=\right.\text { initial compression) } \\&\therefore \mathrm{x}_{\mathrm{i}}=\frac{\mathrm{F}+\mathrm{m}_{1} \mathrm{~g}}{\mathrm{k}} \ldots(1)\end{aligned}$

Let $x_{f}$ be the extension in spring when the lower block just lifts. Also,

$k x_{f}=m_{2} g \text { or } x_{f}=\frac{m_{2} g}{k} \text {...(2) }$

Now from conservation of mechanical energy, increase in gravitational implicit energy of $\mathrm{m}_{1}=$elastic energy of the spring.

$\therefore \mathrm{m}_{1} \mathrm{~g}\left(\mathrm{x}_{\mathrm{i}}+\mathrm{x}_{\mathrm{f}}\right)=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_{\mathrm{i}}^{2}-\mathrm{x}_{\mathrm{f}}^{2}\right)$

Substituting the values of $x_{i}$ and $x_{f}$ from Equations (1) and (2), we get $\mathrm{F}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{g}$

Therefore, the required expression for force is $\mathrm{F}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{g}$.