Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 79
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Andrey seems to say that the solution is simple, but intuitively unclear.
The force must be applied (to the small one) so that the potential energy of the compressed/extended spring is enough to shift the big one. If the force is Kmg, at some point k*x = Kmg (k is the spring's coefficient of elasticity), and the small body can no longer move. The same kx = Kmg will act on the big body, and this will definitely not be enough. So we need more, and not on the epsilon.
We need to apply K(m+delta)g = kx so that kx = K(m+delta)g = KMg.
That is, K(m+delta)g = KMg.
Hence m+delta = M. i.e. delta = M - m.
So the force is K*M*g.
P.S. It was wrong, I corrected it. But it is if applied to small. If applied to large, also there is no way to get less, because you have to shift it all the same.
You have to apply (to the small one) such a force so that the potential energy of the compressed/extended spring is enough to shift the big one. If the force is Kmg, then at some point k*x = Kmg (k is the spring's coefficient of elasticity), and the small body will no longer be able to move further. The same kx = Kmg will act on the big body, and this will definitely not be enough. So we need more, and not on the epsilon.
We need to apply K(m+delta)g = kx so that kx = K(m+delta)g = KMg.
That is, K(m+delta)g = KMg.
Hence m+delta = M. i.e. delta = M - m.
So, the force is equal to K*M*g.
P.S. It was wrong, I corrected it. But it is if applied to small. If applied to large, also there is no way to get less, because you have to shift it all the same.
You're not taking into account that the first box is accelerating.
For example, imagine that we just clicked it and it rolled along its inertia stretching the spring - if the click was strong enough, the second drawer will move, though we do not apply any force to the first drawer at this moment - the moment of shift.
It is the same here: until the first box has a reserve of kinetic energy, which the system pumps into the potential energy of the second box. In layman's terms, the moving box has some inertia, which "helps" the force acting on it to act on the spring and the standing second box.
You also didn't take friction into account.
It is not accelerating. More precisely, the moment the spring and force on the small are equal, they are already in equilibrium (not moving). The spring also strains and prevents it from accelerating.
A snap is a one-step application of a large force (momentum/time of the snap). And we aim to minimise the force. At the moment the big one stalls, the small one stands still as it is balanced by the spring. If it doesn't stand but keeps moving, then the applied force was even greater than KMg.
What is the friction to consider?
P.S. The most convincing proof would be that it doesn't give a shit which box to act on. Then the solution is obvious: we act on the big one.
You don't take into account that the first box is accelerating.
First it accelerates, then it starts to decelerate, then it pushes the second box - if there was enough energy.
In fact, acceleration (kinetic reserve) also depends on the spring stiffness. If the spring is very stiff, for example a steel bar, the acceleration effect will not help, because it tends to zero.
For the second box to move, the spring has to pull it with a force k*M*g. On the other hand, the same force is equal to u*X, where u is the coefficient from Hooke's law (spring stiffness), and X is the distance the first box travelled. Note that throughout this distance it was subjected to friction force k*m*g and force F external to the system. Their total work is equal to (F-k*m*g)*X. The spring tension force is internal to the system and moreover potential (not dissipative), so all its work flows into the potential energy of spring tension. At the moment of detachment this energy according to our conditions is equal to u*(X^2)/2.
So, minimum force F can be obtained from the condition that the total work of external forces must be equal to the potential energy accumulated inside the system. We obtain a system of equations:
k*M*g = u*X
(F-k*m*g)*X = u*(X^2)/2
Substitute u*X from the first equation into the second equation and after reducing X we obtain F = k*(m+M/2)*g.
Which box to influence? To do this, simply note that from (m+M/2)<(M+m/2) follows m<M and vice versa. Conclusion - it is the small box that must be affected.
Which box to influence? To do this, simply note that from (m+M/2)<(M+m/2) follows m<M and vice versa. The conclusion is that it is the small box which has to be affected.
Now connect the boxes with a steel rod (a variant of spring) and try to push it with this formula.
// Hint: it seems to me that Hooke was shot prematurely.
alsu: Подставляем u*X из первого уравнения во второе и после сокращения X получаем F = k*(m+M/2)*g.
We check the special cases - the extreme cases.
Are you implicitly implying that if the boxes are equal, you need 3/2*K*m*g?
Now connect the boxes with a steel bar (a version of a spring) and try to push this formula into place.