What force should be applied to the upper weight of mass m_1 so that the lower weight of mass m_2, connected to the upper one by a spring, breaks off from the floor after this force ceases?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#2.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.3.46.$ What force should be applied to the upper weight of mass $m_1$ so that the lower weight of mass $m_2$, connected to the upper one by a spring, breaks off from the floor after this force ceases?

    </p>

    <h3>Solution</h3>

Forces acting on the upper body in the force-less situation ($F=0$)
In the case where no external force $F$ is applied, we have the equation:
$$N+k\Delta x'=m_2g$$
As $N=0$:
$$\Delta x' = \frac{m_2g}{k}$$
For the case where there is a force $F$
$$\Delta x = \frac{F+m_1g}{k}$$
Law of energy preservation
$$m_1g\left(l-\Delta x\right)+\frac{k\Delta x^2}{2} = m_1g\left(l+\Delta x'\right)+\frac{k{\Delta x'}^2}{2}$$
$$-m_1g\left(\Delta x + \Delta x'\right)=\frac{k}{2}\left({\Delta x'}^2-{\Delta x}^2\right)$$
The expression ${\Delta x'}^2 - {\Delta x}^2$ can be rewritten using the difference of squares:
$$2m_1g=k\Delta x - k\Delta x'$$
Substituting the previously obtained values
$$2m_1g=F+m_1g-m_2g$$
From where the final result could be obtained
$$\boxed{F=m_1g+m_2g}$$

    <h4>Answer</h4>
    <p>
        $$F=(m_1+m_2)g$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      Almaskhan Arsen<br>
    </p>


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