A part of the mass m is detached from the load hanging on the spring of stiffness k. To what height will the remaining part of the cargo then rise?

Savchenko Solutions

Solutions of Savchenko Problems in Physics
knowledge must be free

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

    <h3> Statement </h3>
    <p>
        $2.3.41.$ A part of the mass $m$ is detached from the load hanging on the spring of stiffness $k$. To what height will the remaining part of the cargo then rise?

    <h3>Solution</h3>
    <p>
        Since there is no interaction with external factors other than breakaway, all the work goes into changing the length of the spring. $E_k=0$!

Small change in the pontential energy of gravity $$dA = mg \\, dx$$ Small change in the pontential energy of the spring $$dE = kx \\, dx$$ Law of conservation of energy $$dA = dE$$ $$mg \\, dx = kx \\, dx$$ $$mg \int dx = k \int x \\, dx$$ From where $$mg \\, x = \frac{kx^2}{2} \Rightarrow \boxed{x = \frac{2mg}{k}}$$

    <h4>Answer</h4>
    <p>
        $$x = \frac{2mg}{k}$$
    </p>
    
    <p style="text-align: right; font-style: italic; font-size: 14;">   
      Almaskhan Arsen<br>
    </p>


<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
    </p>
</footer>