At the ends of a long thread, weights of mass m each are suspended. The thread is spanned over two light little blocks located at a distance of 2l from each other. A 2m weight is attached to it in the middle between the blocks, and the system starts moving. Find the speed of loads after a sufficiently long period of time has elapsed.

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    <h3> Statement </h3>
    <p>
        $2.4.9^*.$ At the ends of a long thread, weights of mass $m$ each are suspended. The thread is spanned over two light little blocks located at a distance of $2l$ from each other. A $2m$ weight is attached to it in the middle between the blocks, and the system starts moving. Find the speed of loads after a sufficiently long period of time has elapsed.

    <h3>Solution</h3>
    <p>

From the geometry of the drawing, a trigonometric expression can be obtained by
$$\tan\varphi=\frac{l}{h}\quad(1)$$
$$\sin\varphi=\frac{l}{\sqrt{l^2+h^2}};\quad\cos\varphi=\frac{h}{\sqrt{l^2+h^2}}\quad(2)$$
Over the long period of time the $2m$ weight would go down infinitely ($h\to \infty$)
From $(1)$, we obtain that as $h$ increases $\tan\varphi$ decreases
$$\tan\varphi \to 0;\quad\varphi \to 0$$
For the small angle $\varphi$ we could use an approximation to the second order of magnitude
$$\sin\varphi \approx\tan\varphi\approx\varphi;\quad\cos\varphi\approx 1$$
From the geometry of the figure, the length of the thread
$$\sqrt{l^2+h^2}=l+\Delta x$$
by the Pythagorean theorem, the change in the bottom of the thread connecting $2m$
$$\Delta x = \sqrt{l^2+h^2}-l$$
Similarly, from the geometry of the figure

Considering $\sin\varphi\approx\tan\varphi$ and $h\gg l$, we obtain
$$\sqrt{l^2+h^2}\approx h\Rightarrow \boxed{l=h-\Delta x}\quad(2)$$
$L_1,~L_2$ — Length of threads at the initial moment

Conservation of the mechanical energy $(dE_p+dE_k=0)$

Two weights with mass $m$ went up to $\Delta x$ and the $2m$ went down $h$, making change in potential energy
$$dE_p=-2mgh+2\cdot mg\Delta x$$
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
$$dE_p+dE_k=0\Leftrightarrow -2mgh+2\cdot mg\Delta x+2\frac{mv_1^2}{2}+\frac{2m\cdot v_2^2}{2}=0$$
$$v_1^2+v_2^2=2g\left(h-\Delta x\right)\quad(3)$$
By the Newton's second law for the $m$ weight
$$ma=T-mg$$
Similarly, for the $2m$ weight
$$2ma^*=2mg-2T\cos\varphi\approx2mg-2T\Leftrightarrow$$
$$\Leftrightarrow ma^*=mg-T$$
As they are connected by the same inextensible thread, their acceleration will be equal
$$\left|a\right|=\left|a^*\right|\Rightarrow v=v^*$$
Given that the system is being run from a stationary state $(v_0=v_0^*=0)$, let's substitute into $(3)$
$$2v^2=2g\left(h-\Delta x\right)$$
Taking into consideration $(2)$
$$v^2=gl\Rightarrow \boxed{v=\sqrt{gl}}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$v=\sqrt{gl}$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      Almaskhan Arsen<br>
    </p>


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