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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>
+ $2.3.36^*.$ The path length of a particle of mass $m$ is proportional to its initial momentum if the force braking the particle is proportional to its velocity (see the previous problem). Make sure of this and for a given $\alpha (l = \alpha p)$, find the work of the braking force on the path $x$ for a particle with mass $m$ and initial momentum $p$.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ Let's make sure that $F \sim v$!
+$$dA = Fdx \\$$
+$$dE = \frac{P dP}{m} \\$$
+Hence
+$$\frac{mv^2}{2} = \frac{P^2}{2m} \\$$
+Find the small increment of energy $E$
+$$d(E) = d\left(\frac{P^2}{2m}\right) = \frac{P dP}{m} = \frac{PdP}{m} \\$$
+From the law of conservation of energy, this energy has completely gone into doing work $A$:
+$$dA = dE \\$$
+$$Fdx = \frac{P dP}{m} \\$$
+$$dx = \alpha dP \Rightarrow l = \alpha P \\$$
+$$\frac{dP}{dx} = \frac{1}{\alpha} \\$$
+$$F = \frac{\alpha P}{m \alpha^2} = \frac{v P}{\alpha} \Rightarrow F \sim v$$
+That's what we had to prove!
+</p><p>
+The work of force $F$ on a small interval $dx$ is found as
+$$dA = Fdx = \frac{P}{m \alpha^2}x dx - \frac{P}{m \alpha} dx \\$$
+And the total work could be found as a sum of all $dA$ all along the way
+$$\int_0^A dA = \int_0^x \frac{x dx}{m \alpha^2} - \int_0^x \frac{P dx}{m \alpha} \\$$
+$$\boxed{A = \frac{x^2}{2m \alpha^2} - \frac{Px}{m \alpha}}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$A = \frac{x^2}{2m \alpha^2} - \frac{Px}{m \alpha}$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Almaskhan Arsen<br>
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