$2.3.36^*.$ 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 $p$.
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 $A$: $$dA = dE \\$$ $$Fdx = \frac{p dp}{m} \\$$ $$dx = \alpha dp \Rightarrow l = \alpha p \\$$ $$\frac{dp}{dx} = \frac{1}{\alpha} \\$$ $$F = \frac{p}{m \alpha} = \frac{v}{\alpha} \Rightarrow \boxed{F \sim v}$$ 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 $$\int_0^A dA = \int_0^x \frac{x dx}{m \alpha^2} - \int_0^x \frac{p dx}{m \alpha} \\$$ $$\boxed{A = \frac{x^2}{2m \alpha^2} - \frac{px}{m \alpha}}$$
$$A = \frac{x^2}{2m \alpha^2} - \frac{Px}{m \alpha}$$
Almaskhan Arsen