Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 84

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Mathemat:


P.S. It seems that this is also a forum virus like the problem of the plane.

There !

Which means we have to vote.

And since the moderator has k=10 when voting, we win.

 
Mathemat:

Yes, technically there are no errors to be seen. But who says that only the big box will be touched? The small one will get it too, the spring doesn't give a shit where to act...

P.S. It seems that it is also a forum virus like a problem about an aeroplane.

At the beginning, the spring does not act on the small one at all, so we just start pulling it (the calculated force is enough to overcome the rest friction). The further we pull, the more the spring prevents this from happening. At the end it acts on the small one with force k*M*g, the same as on the big one, friction force k*m*g acts in the same direction, so the equilibrium will already be directed backwards. This means that by the time the second box moves, the first box will already be slowing down for a while (I suspect it will just stop).
 
Mischek: And since the moderator has k=10 in the voting, we win.
Am I correct in assuming that k is the coefficient of friction?
 
TheXpert:

No. The process will stall. (sort of).

It seems that it won't stall, for this reason: if we were able to move the centre of mass of the system once by the force F, then we can move it any number of times more.
 
Mathemat:
Am I correct in assuming that k is the coefficient of friction here?
What about what? (whatever your anatomy as moderators)
 
alsu:
In the beginning, the spring does not act on the small one at all, so we just start pulling it (the calculated force is enough to overcome the friction of rest). The further you pull, the more the spring prevents it from doing so. At the end it acts on the small one with force k*M*g, the same as on the big one, friction force k*m*g acts in the same direction, so the equilibrium will already be pointing backwards. This means that by the time the second box moves, the first box will already be slowing down for a while (I suspect it will stop).
This is all logical, but it is also logical that the accumulated kinetic energy will depend on the time elapsed from the start of the first body's movement until the second body moves (because the force is constant). Therefore: the softer the spring, the less force is needed.
 
alsu:
It doesn't seem to stall, for this reason: if we were able to move the centre of mass of the system once by the force F, we can move it any number of times more.

This is only without friction.

(?)

 
Mathemat:
Am I correct in assuming that k is the coefficient of friction?

)))

No, it's just that your vote equals ten.

Looks like we're done with friction today.

 
MetaDriver:
This makes sense, but it also makes sense that the stored kinetic energy will depend on the time elapsed from the start of the first body's movement until the second body moves (because the force is constant). Therefore, the softer the spring, the less force is needed.
But it will take longer for the spring to pull the second box with the right force
 
MetaDriver:

It's only without the friction.

(?)

The first time was with friction, so can it be the second time?
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