Alright, first time I'm answering a physics question on this forum, so bare with me
(Oh, an I don't know if you have a formula editor built into this page, sry for not using it now if there is one.)
Start with the basing GCSE formula for distance cover while having constant acceleration:
s=(at^2)/2
We already have s and t given, what is a? Well, you have friction acting upon the body, which, in this case, is simply weight times the friction coefficient.
a = F/m = fmg/m = fg
Therefore:
s = (fgt^2)/2
Rearranging and plugging in the numbers will give you f = 0.05 [dimensionless]. Now ask yourself what actually happened. The object travelled exactly 1m and stopped, therefore if f = 0.05 then it will stop at the opposite side of the table, not fall. You can also find that the initial velocity of the object must have been exactly 1 m/s (you can calculated this yourself). So, if you want the object to fall, you have to decrease the coefficient of friction below 0.05 AND decrease the initial velocity to meet your conditions. The other extreme situation is when f=0 and v_0 = 0.5m/s. You could probably figure out the relationship between f and v_0, but I'll leave that to you.
Dunno, how this proves/disproves whether the object has wheels, but I hope this helps