To maintain constant velocity $v$, the traction force $F_T$ must provide the impulse to accelerate the newly added mass from zero horizontal velocity to $v$. In a time interval $\Delta t$, a mass $\Delta m = \lambda \Delta t$ is added. Its horizontal momentum changes by $\Delta p = (\Delta m)v$. The impulse required is $F_T \Delta t = \Delta p$. [Q1] The required force is $F_T = \frac{\Delta p}{\Delta t} = \frac{(\Delta m)v}{\Delta t} = (\frac{\Delta m}{\Delta t})v = \lambda v$. [Q2] The work done in time $\Delta t=1$ s is $W = F_T \cdot d = F_T (v \Delta t)$. For $\Delta t=1$ s, $W = F_T v = (\lambda v)v = \lambda v^2$. [Q3] The mass of coal added in $\Delta t=1$ s is $\Delta m = \lambda$. Its kinetic energy gain is $\Delta K = \frac{1}{2}\Delta m v^2 = \frac{1}{2}\lambda v^2$. [Q4] Comparing $W = \lambda v^2$ and $\Delta K = \frac{1}{2}\lambda v^2$, we see that $W = 2\Delta K$. They are not equal. Half the work is converted into thermal energy during the inelastic process of the coal coming to rest relative to the car.