Added 2.7.20 solution

astrosander edited
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revision #8762 parent #8761 GitHub 534beb8 ← older newer →
en/2.4.9.md +2 −2
@@ -95,11 +95,11 @@
$L_1,~L_2$ — Length of threads at the initial moment</p><p>
Conservation of the mechanical energy $(dE_p+dE_k=0)$</p><p>
Two weights with mass $m$ went up to $\Delta x$ and the $2m$ went down $h$, making change in potential energy
−$$dE_p=2mgh-2\cdot mg\Delta x$$
+$$dE_p=-2mgh+2\cdot mg\Delta x$$
In the meantime, the velocities of $m$ and $2m$ bodies become $v_1$ and $v_2$, respectively, making total kinetic energy of the system
$$dE_k = 2\frac{mv_1^2}{2}+\frac{2m\cdot v_2^2}{2}$$
Substituing into equations of conservation of the mechanical energy
−$$dE_p+dE_k=0\Leftrightarrow 2mgh-2\cdot mg\Delta x+2\frac{mv_1^2}{2}+\frac{2m\cdot v_2^2}{2}=0$$
+$$dE_p+dE_k=0\Leftrightarrow -2mgh+2\cdot mg\Delta x+2\frac{mv_1^2}{2}+\frac{2m\cdot v_2^2}{2}=0$$
$$v_1^2+v_2^2=2g\left(h-\Delta x\right)\quad(3)$$
By the Newton's second law for the $m$ weight
$$ma=T-mg$$
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