 Solution

The inclined parcel has tension $T$ up the slope, weight $mg$ vertically downward, normal reaction $N=mg\cos \theta$ and friction $\mu mg\cos \theta$ down the slope. The hanging mass has weight downward and tension upward. The ideal string makes the two acceleration magnitudes equal.

The velocity slope is $a=2.4/1.5=1.60\mathrm{m}{\mathrm{s}}^{-2}$, up the track for the parcel and downward for the hanging mass. The free-body diagrams show gravity itself as one vector; its slope-parallel and normal components are used in the equations rather than counted as additional forces.