Volcanic eruption force
For visualization purposes, if the $35,000$ tonne rock was composed of something similar to basalt, with a density of $3 g/cm^3$ (also $3 t/m^3$), it would be a sphere with a diameter of $28.1$ m, or a cube with sides of $22.7$ m. 

Using a ballistic trajectory calculator, to move anything $2000$ km down range, the object needs an initial velocity of $4430$ m/s ($15,948$ km/h, or Mach $12.9$ - hyper sonic speed), assuming a projection angle of $45$ degrees to the horizon. The rock would reach height of $500$ km.

The kinetic energy of such an object would be $3.434(10)^{14}$ J, using the equation $E_k = 0.5mv^2$.

When Mount St. Helens erupted in 1980, it released $24$ megatons of thermal energy. Using a converter, this is equivalent to $1.004(10)^{17}$ J. This is 292 times the energy required to move a $35,000$ tonne rock $2000$ km.

Now, the energy released by Mount St. Helens was the overall energy released, not the energy given to one lump of rock and Mount St. Helens was not a supervolcano. So, could a supervolcano shoot a $35,000$ tonne rock $2000$ km away? Potentially, but unlikely. Energy always takes the path of least resistance so instead of moving a $35,000$ tonne rock $2000$ km away it will more likely move significantly more material that is in the form of dust or ash.