Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 85

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alsu:
But longer will have to be pulled so that the spring can pull with the right force the second drawer
Longer isn't forbidden. I like longer.
 
alsu:
The first time was with friction, so can you do the second?
Only if the vector is switched. Then you can jerk both in place.
 

Alexei, I don't understand the contradiction between your reasoning

alsu: Для того, чтобы второй ящик поехал, необходимо, чтобы пружина потянула его с силой k*M*g. С другой стороны, та же сила равна u*X, где u - коэффициент из закона Гука (жесткость пружины), а X - расстояние, который прошел первый ящик. Отметим, что на протяжении этого пути на него действовали внешние по отношению к системе сила трения k*m*g и сила F. Их суммарная работа равна (F-k*m*g)*X. Сила натяжения пружины - внутренняя для данной системы и , кроме того, потенциальная (не диссипативная), поэтому вся ее работа перетекает в потенциальную энергию натяжения пружины. В момент отрыва эта энергия согласно нашим условиям равна u*(X^2)/2.

So, the minimum force F can be obtained from the condition that the total work of external forces must be equal to the accumulated potential energy inside the system. We obtain a system of equations:

k*M*g = u*X

(F-k*m*g)*X = u*(X^2)/2

Let's substitute u*X from the first equation into the second one and after reducing X we get F = k*(m+M/2)*g.

and by mine:

M: OK, let's take friction into account. Apply K(m+delta)g. The acceleration starts, the spring compresses/extends.

The balance of forces is such that only K*delta*g acts on the spring due to friction energy consumption, which will charge it and push the big body when the spring fully balances the small one and it stops.

It turns out that you need K(m+M)g. Again it turns out that it doesn't matter which body to push.

But yours has energy and mine doesn't need it... Where the hell is my mistake?

 
I remember a phrase circulating among creationists that if you shake watch parts in a box for a long time, you won't be able to assemble the watch. It was around until the dude wrote an emulator and showed that not only do watches gather at random shaking, but they also evolve, with more adapted ones knocking out their less adapted ancestors)))).
 
alsu:
It seems that it will not stall, for the following reason: if we were able to displace the centre of mass of the system once by F, we can do it any number of times more.
There is a difference between the initial situation and the situation after the influence of F . Despite the displacement of the centre of mass, we obtained compression of the springs
 
alsu:

I remember, among creationists there was a phrase that if you shake the watch parts in a box for a long time, you can't assemble the watch. It was around for a while, until the guy wrote a suitable emulator, and did not show evidently that the watch is not only collected at random shaking, but also evolves, with more fit ones knocking out their less fit ancestors)))).

!!

this is ten!!!!!!

 

Mathemat:

The balance of forces is such that due to frictional energy expenditure only K*delta*g acts on the spring, which will both charge it and push the big body when the spring fully balances the small one and it stops.

Where the hell is my mistake?

There's a bigger force acting on the spring than you wrote - we're still pulling the box with a force F, and that force minus the friction is transferred to both the spring and the second box
 
Mischek:
Nope, there is a difference between the initial situation and after F. Despite the displacement of the centre of mass, we got compression of the springs
Okay. Let's say that at the moment the second box comes off, we let go of the force. Where will the centre of mass stop? It seems to me that not in the initial position (it will have to go back there!). Now let's return the force to the problem. All other things being equal, we got an addition in the form of a force acting in the direction of motion, which in no way can move the system backwards.
 
alsu:
There is a larger force acting on the spring than you wrote - we are also pulling the box with force F, and this force minus friction is transferred to both the spring and the second box

If we have applied K(m+delta)g to a small one, only K*delta*g acts on the spring, as Kmg goes to friction. Is this wrong?

I don't understand how you can move a body of mass (M+m) by applying less than K(M+m)g.

 
Mathemat:

I don't understand how you can move a body of mass (M+m) by applying less than K(M+m)g.

In a hurry.
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