Is it true that open clusters have $E > 0$ and globular clusters have $E < 0$? I just need to confirm or refute something I'm believing since a very long time, maybe without real evidence.

Can we distinguish stellar clusters by their total mechanical energy ?

Is it true that all "open" star clusters have a positive energy (i.e. more kinetic energy that potential energy), while all globular clusters are "closed" systems with negative total energy (i.e less kinetic energy than total potential energy) ?

So for example, the Peiades M45 are "open" : they have $E &gt; 0$, while M4, M53 and M80 (globulars) have $E &lt; 0$ ?  Is this true ?
Stars in an open star cluster are - according to wikipedia - "loosly bound by mutual gravity". Whereas a globular cluster is more dense and will be "less loosly" bound - to use the same terminology. 

Since the stars in the open star cluster still are gravitationally bound, the total energy $E$ must be less than zero. This criterium that $E&lt;0$ is always correct for (gravitationally or otherwise) bound systems for exactly the reason you stated, the kinetic energy is 'less positive' than the potential is negative. Otherwise the particles (stars in this case) would break free from the gravitational interactions with the other particles - and that's the complete opposite of a bound system.

However, the fate of most open star clusters is that it will fall apart. This can be due to 


mass loss of the cluster or,
internal processes like a slingshot due to close encounters between constituent stars.


From the get-go these open clusters are quite unstable, so $E$ will be close to zero, but still negative ofcourse because it's a bound system. But it is instable enough for the following to happen. Young, bright stars radiate a lot of energy, and this means that the radiation pressure on the surronding cloud from which these stars are born is high. The light from these young stars can blow large amounts of mass in the form of this gas and dust out of the system - which has an influence on the total energy of the system! Because the mass of the system drops, the total energy rises (read: becomes less negative) - wich makes it easier for the constituent stars to reach escape velocity. If you don't like thinking in terms of energy, the escape velocity is $v_e = \sqrt{2GM/r}$, so $v_e \propto M $. If $M$ drops, so does the escape velocity. 

The second bullet speaks for itself, I think.

You can deduce that $E$ has to be negative from a conservation viewpoint. When the stars in the cluster are born from the same cloud, this cloud already is gravitationally bound, so $E&lt;0$, and unless some energy is lost at some point, the cloud will stay bound. But like I said and as you said, most of these clusters don't stay bound for very long, and eventually will evaporate.