Avals: I think the point here is that if a force is applied to a small box, it will start moving with acceleration and will have variable momentum and kinetic energy. Therefore you need less force than (M+m)Kg but more than mKg
OK, mKg + eps. One has to move one box anyway. Let it be a small one.
TheXpert: On a smooth surface there are two boxes connected by a spring .
Their masses are M and m, (M > m) the coefficient of friction is K. A constant force F acts on one of the boxes.
What minimum force F is needed and on which box to apply it to make both boxes move.
Apparently, it is required to find the minimumconstant force that needs to be applied for some time. The condition is not formulated quite correctly.
Everyone who is solving - it is necessary to take into account that at the moment of the second box collapse the first box has already gained some speed (it moved accelerated, because the spring reaction force was not constant, but grew from 0), so F needed for collapse will be less than in statics. I kind of got k(m+M/2)g, I am not writing the solution yet.
27.4. Потенциальная энергия деформированной пружины Деформированная (например, растянутая) пружина способна совершить работу (рис. 155). рис. 155 Действительно, если к растянутой пружине прикрепить некоторое тело, то пружина будет действовать на него с некоторой силой, под действием которой тело начнет смещаться. Следовательно, будет...
On a smooth surface there are two boxes connected by a spring
.
A constant force F acts on one of the boxes.
What minimum force F is needed and on which box to apply it to make both boxes move.
Apparently, it is required to find the minimum constant force that needs to be applied for some time. The condition is not formulated quite correctly.
Everyone who is solving - it is necessary to take into account that at the moment of the second box collapse the first box has already gained some speed (it moved accelerated, because the spring reaction force was not constant, but grew from 0), so F needed for collapse will be less than in statics. I kind of got k(m+M/2)g, I am not writing the solution yet.
alsu:
I kind of got k(m+M/2)g, I don't write the solution yet.
Well, or k(M+m/2)g, whichever is smaller.
Ps But the former is smaller if m<M))
It's probably about the work that needs to be done.
Yes
The challenge for loonies is to solve the same problem if you can change the vector of force arbitrarily.
about that - it doesn't seem to matter. Since the work done does not depend on the trajectory of the body http://fizportal.ru/physics-book-27-4