Updated en 2.2.37, 2.2.44, 2.2.36, 2.2.39

astrosander edited
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revision #2514 parent #2504 GitHub f9d1737 ← older newer →
en/2.3.36.md +2 −2
@@ -61,11 +61,11 @@
$$Fdx = \frac{p dp}{m} \\$$
$$dx = \alpha dp \Rightarrow l = \alpha p \\$$
$$\frac{dp}{dx} = \frac{1}{\alpha} \\$$
−$$F = \frac{\Delta p}{m \alpha^2} = \frac{\Delta v}{\alpha} \Rightarrow \boxed{F \sim v}$$
+$$F = \frac{p}{m \alpha} = \frac{v}{\alpha} \Rightarrow \boxed{F \sim v}$$
That's what we had to prove!
</p><p>
The work of force $F$ on a small interval $dx$ is found as
−$$dA = Fdx = \frac{p^*-p}{m \alpha} dx - \frac{p}{m \alpha} dx $$
+$$dA = Fdx = \frac{p^*-p}{m \alpha} dx$$
$$dA = \frac{x}{m \alpha^2} dx - \frac{p}{m \alpha} dx$$
And the total work could be found as a sum of all $dA$ all along the way
$$\int_0^A dA = \int_0^x \frac{x dx}{m \alpha^2} - \int_0^x \frac{p dx}{m \alpha} \\$$
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