The body of mass m is attached to two consecutive springs of stiffness k_1 and k_2. A constant force F is applied to the free end of the spring chain. What is the total elongation of the springs, if the vibrations have already stopped?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

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    <h3> Statement </h3>
    <p>
        $2.1.17.$ The body of mass $m$ is attached to two consecutive springs of stiffness $k_1$ and $k_2$. A constant force $F$ is applied to the free end of the spring chain. What is the total elongation of the springs, if the vibrations have already stopped?

    <h3>Solution</h3>
    <p>
        <center>

1. The body can be considered as free if the connections are replaced by their reactions. The friction force in this case is caused by the action of the magnetic force, i.e. $$F_{fr} = \mu F_m$$ 2. The condition of mass equilibrium, in this case, will take place when the magnitudes of gravity and friction are equal. $$m_0 g = \mu F_m\Rightarrow \boxed{F_m = \frac{gm_0}{\mu}}$$ 3. When the body starts moving at $m > m_0$, then the equation of Newton's second law becomes valid, which in projection onto the $y$ axis will be written as follows $$mg - m_0 g = ma$$

$$\boxed{a = g \frac{m - m_0}{m}}$$

    <h4>Answer</h4>
    <p>
                 $$F = m_0g/\mu ;$$
     $$a = g(m − m_0)/m$$
    </p>


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