Added 2.2.44; Updated 2.3.36 2.3.44

astrosander edited
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revision #2504 parent #2502 GitHub 34e8c2b ← older newer →
en/2.3.36.md +3 −2
@@ -61,11 +61,12 @@
$$Fdx = \frac{p dp}{m} \\$$
$$dx = \alpha dp \Rightarrow l = \alpha p \\$$
$$\frac{dp}{dx} = \frac{1}{\alpha} \\$$
−$$F = \frac{\alpha p}{m \alpha^2} = \frac{v p}{\alpha} \Rightarrow F \sim v$$
+$$F = \frac{\Delta p}{m \alpha^2} = \frac{\Delta v}{\alpha} \Rightarrow \boxed{F \sim v}$$
That's what we had to prove!
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The work of force $F$ on a small interval $dx$ is found as
−$$dA = Fdx = \frac{p}{m \alpha^2}x dx - \frac{p}{m \alpha} dx \\$$
+$$dA = Fdx = \frac{p^*-p}{m \alpha} dx - \frac{p}{m \alpha} dx $$
+$$dA = \frac{x}{m \alpha^2} dx - \frac{p}{m \alpha} dx$$
And the total work could be found as a sum of all $dA$ all along the way
$$\int_0^A dA = \int_0^x \frac{x dx}{m \alpha^2} - \int_0^x \frac{p dx}{m \alpha} \\$$
$$\boxed{A = \frac{x^2}{2m \alpha^2} - \frac{px}{m \alpha}}$$
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