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 lpha (l = lpha p), find the work of the braking force on the path x for a particle with mass m and initial momentum p.

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    <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} $$Hence d(E) = d\left(\frac{p^2}{2m}\right) = \frac{p dp}{m} = \frac{pdp}{m}$
From the law of conservation of energy, this energy has completely gone into doing work $A$:
$$dA = dE $$$Fdx = \frac{p dp}{m} $$\frac{dp}{dx} = \frac{1}{\alpha}$
$$F = \frac{p}{m \alpha} = \frac{v}{\alpha} \Rightarrow \boxed{F \sim v}$$
That's what we had to prove!

The work of force $F$ on a small interval $dx$ is found as $$dA = Fdx = \frac{p^*-p}{m \alpha} dx$$ $$dA = \frac{x}{m \alpha^2} dx - \frac{p}{m \alpha} dx$$ 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}}$$

    <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>
    </p>


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