Added en 2.3.35 2.3.39 2.3.41

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+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <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?
+
+</p>
+<center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" width="100" />
+ <figcaption>
+ For problem $2.3.41$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <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. $E_k=0$!
+</p><p>
+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
+$$
+mg \, x = \frac{kx^2}{2} \Rightarrow \boxed{x = \frac{2mg}{k}}
+$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$x = \frac{2mg}{k}$$
+ </p>
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+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Almaskhan Arsen<br>
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