 Solution

Let $\dot{M}=500\mathrm{kg}{\mathrm{s}}^{-1}$ and $V=10\mathrm{m}{\mathrm{s}}^{-1}$. Incoming material has zero horizontal momentum, while incorporation requires horizontal momentum $VdM$. Thus the engine supplies force $F=V\dot{M}$, and its power isNo duration or loading distance is given, so a single total work cannot be assigned. Over an interval $t$ in seconds,The remaining $25,000t\mathrm{J}$ becomes internal energy in the inelastic collision that brings the falling material to the wagon's horizontal velocity. Vertical falling kinetic energy, if any, is a separate contribution to dissipation; it does not alter this horizontal work balance. The source alternates between stone and coal; the calculation depends only on the given loading rate.