Constant-burn rocket with ground friction ($q(t)\dot{v}=su-\mu gq(t)$)

For fuel mass rate $s>0$ and backward relative exhaust speed $u$, a rocket sliding on a horizontal plane has remaining mass $q(t)$ and satisfies the displayed rocket equation with kinetic friction. Integration over a sliding interval gives $v(t)-v(t_0)=u\log [q(t_0)/q(t)]-\mu g(t-t_0)$. From initial rest, immediate forward sliding additionally requires thrust to overcome static friction. If both friction coefficients equal $\mu$, motion starts only when $su>\mu gq(t)$. An algebraically negative sliding speed is not the physical solution at rest.