Fixed en 2.3.36 solution

astrosander edited
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revision #2502 parent #2495 GitHub 605f737 ← older newer →
en/2.3.36.md +10 −10
@@ -51,24 +51,24 @@
<p>
Let's make sure that $F \sim v$!
$$dA = Fdx \\$$
−$$dE = \frac{P dP}{m} \\$$
+$$dE = \frac{p dp}{m} \\$$
Hence
−$$\frac{mv^2}{2} = \frac{P^2}{2m} \\$$
+$$\frac{mv^2}{2} = \frac{p^2}{2m} \\$$
Find the small increment of energy $E$
−$$d(E) = d\left(\frac{P^2}{2m}\right) = \frac{P dP}{m} = \frac{PdP}{m} \\$$
+$$d(E) = d\left(\frac{p^2}{2m}\right) = \frac{p dp}{m} = \frac{pdp}{m} \\$$
From the law of conservation of energy, this energy has completely gone into doing work $A$:
$$dA = dE \\$$
−$$Fdx = \frac{P dP}{m} \\$$
−$$dx = \alpha dP \Rightarrow l = \alpha P \\$$
−$$\frac{dP}{dx} = \frac{1}{\alpha} \\$$
−$$F = \frac{\alpha P}{m \alpha^2} = \frac{v P}{\alpha} \Rightarrow F \sim v$$
+$$Fdx = \frac{p dp}{m} \\$$
+$$dx = \alpha dp \Rightarrow l = \alpha p \\$$
+$$\frac{dp}{dx} = \frac{1}{\alpha} \\$$
+$$F = \frac{\alpha p}{m \alpha^2} = \frac{v p}{\alpha} \Rightarrow F \sim v$$
That's what we had to prove!
</p><p>
The work of force $F$ on a small interval $dx$ is found as
−$$dA = Fdx = \frac{P}{m \alpha^2}x dx - \frac{P}{m \alpha} dx \\$$
+$$dA = Fdx = \frac{p}{m \alpha^2}x dx - \frac{p}{m \alpha} dx \\$$
And the total work could be found as a sum of all $dA$ all along the way
−$$\int_0^A dA = \int_0^x \frac{x dx}{m \alpha^2} - \int_0^x \frac{P dx}{m \alpha} \\$$
−$$\boxed{A = \frac{x^2}{2m \alpha^2} - \frac{Px}{m \alpha}}$$
+$$\int_0^A dA = \int_0^x \frac{x dx}{m \alpha^2} - \int_0^x \frac{p dx}{m \alpha} \\$$
+$$\boxed{A = \frac{x^2}{2m \alpha^2} - \frac{px}{m \alpha}}$$
</p>
<h4>Answer</h4>
unchanged lines 20