= Solution
Let $\dot M=500\,\mathrm{kg\,s^{-1}}$ and $V=10\,\mathrm{m\,s^{-1}}$. Incoming material has zero horizontal <momentum>, while incorporation requires horizontal <momentum> $V\,dM$. Thus the engine supplies <force> $F=V\dot M$, and its <power> is
$$
\boxed{F=5000\,\mathrm N,\qquad P=FV=\dot M V^2=50\,\mathrm{kW}.}
$$
No duration or loading distance is given, so a single total <work> cannot be assigned. Over an interval $t$ in seconds,
$$
\boxed{W=50{,}000t\,\mathrm J,\qquad \Delta K_{\rm horizontal}=\tfrac12\dot M V^2t=25{,}000t\,\mathrm J.}
$$
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.
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