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Solution:

Let us write the fundamental equation of dynamics for all the three blocks in terms of projections, having taken the positive direction of x and y axes as shown in Fig; and using the fact that kinematical relation between the accelerations is such that the blocks move with same value acceleration (say w)

 \( {{m}_{0}}g-{{T}_{1}}={{m}_{0}}w\,\,\,\,\,(1) \)

 \( {{T}_{1}}-{{T}_{2}}-k{{m}_{1}}g={{m}_{1}}w\,\,\,\,\,\,\,\,(2) \)

And  \( {{T}_{2}}-k{{m}_{2}}g={{m}_{2}}w\,\,\,\,\,\,\,(3) \)

The simultaneous solution of Eqs. (1), (2) and (3) yields,  \( w=g\frac{[{{m}_{0}}-k({{m}_{1}}+{{m}_{2}})]}{{{m}_{0}}+{{m}_{1}}+{{m}_{2}}} \) and \( {{T}_{2}}=\frac{(1+k){{m}_{0}}}{{{m}_{0}}+{{m}_{1}}+{{m}_{2}}}{{m}_{2}}g \).

As the block m₀ moves down with acceleration w, so  in vector form  \( \vec{w}=\frac{[{{m}_{0}}-k({{m}_{1}}+{{m}_{2}})]}{{{m}_{0}}+{{m}_{1}}+{{m}_{2}}}\vec{g} \).