Does relativistic mass have weight? If an object was sliding on an infinitely long friction-less floor on Earth with relativistic speeds (ignoring air resistance), would it exert more vertical weight force on the floor than when it's at rest?
 A: My previous answer proved to be wrong. Energy density ("relativistic mass") does contribute to gravity - and the fact that the object is moving at relativistic speeds does affect the space-time around it.
There is an interesting document that explains the problem in further context.  
Besides, when we think about it, if energy density didn't contribute to the force of gravity, photons, mass of which consists of their energy density, would not be affected by gravity. But we have provided many times, that the lightwaves - thus photons - are affected by gravity.
A: "Does gravity depend on relativistic mass or rest mass?" is a rather interesting question -- Einstein's initial approach was to say "the relativistic mass", and this was the pre-general relativistic answer, but this is not satisfactory, since the relativistic mass is only one component of the energy-momentum vector (and it would actually sound more reasonable to say it depends on the rest mass).
This is the theoretical motivation for general relativity, in which gravity depends on the stress-energy-momentum tensor, and only the time-time component of gravity, $\Gamma^i{}_{00}$, depends on the relativistic mass, whose density is also the time-time component of the stress-energy tensor, and the other components depend on on the other components of the stress-energy tensor by the Einstein field equations.
A: First off, your question is phrased in terms of relativistic mass, which is an obsolete concept. But anyway, that's a side issue.
The question can be posed in terms of either the earth's force on the puck or the puck's force on the earth. We expect these to be equal because of conservation of momentum.
In general relativity, the source of gravitational fields is not the mass or the mass-energy but the stress-energy tensor, which includes pieces representing pressure, for example. The puck has some stress-energy tensor, and this stress-energy tensor is changed a lot by the puck's highly relativistic motion. Therefore the puck's own gravitational field is definitely changed by the fact of its motion. However, the change is not simply a scaling up of its normal gravitational field. The field will also be distorted rather than spherically symmetric. Yes, the effect is probably to increase its force on the earth. The earth therefore makes an increased force on the puck.
Here is a similar example that shows that you can't just naively use $E=mc^2$ to calculate gravitational forces. Two beams of light moving parallel to each other experience no gravitational interaction, while antiparallel beams do. See Tolman, R.C., Ehrenfest, P., and Podolsky, B. Phys. Rev. (1931) 37, 602, http://authors.library.caltech.edu/1544/
A: Assuming the frictionless floor is locally horizontal, orthogonal to the radial direction and ignoring the effects of the spinning earth, the force on the floor would actually be zero when you reach the velocity of $v=\sqrt{GM/r}$­ where r is the radial distance to the center of the earth. This is because the sliding object would now be in orbit.
I suspect the force on the sliding floor generally could be written as:
$$F=-\frac{mGM}{r^2}\hat{r}+m\frac{v^2}{r}\hat{r}$$
If you have pure non-radial motion. This would become zero long before relativistic effects become noticable. I guess a totally correct answer depends a bit on if the floor "bends with the Earth" so it is always orthogonal to the radial direction. The answer here is correct if the floor "bends with the Earth".
