A cart loaded with sand moves along a horizontal plane due to a constant force F coinciding in direction with the cart’s velocity vector. In the process, sand spills through a hole in the bottom with a constant velocity $\mu$ kg / s . Find the acceleration of the cart at the moment t, if at the initial moment t = 0, the cart with the loaded sand has a mass m₀ and its velocity was equal to zero. The friction is to be neglected.

In this problem, the sand spills through a hole in the bottom of the cart. Hence, the relative velocity of the sand ${\mathrm{V}}_{\mathrm{r}}$​ will be zero because it will acquire the same velocity as that of the cart at the moment.
${\mathrm{v}}_{\mathrm{r}} = 0$

Thus, ${\mathrm{F}}_{\mathrm{t}} = 0$ (as ${\mathrm{F}}_{\mathrm{t}} = {\mathrm{v}}_{\mathrm{r}} \frac{\mathrm{dm}}{\mathrm{dt}}$​)
and the net force will be F only. 
∴${\mathrm{F}}_{\mathrm{net}}$=F
or m($\frac{\mathrm{dv}}{\mathrm{dt}}$)F............(i)
But here m = ${\mathrm{m}}_0$​ − μt

∴ ($m_0$−μt)$\raisebox{1ex}{$\mathrm{dv}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dt}$}\right.$​  = F
or ${\int }_0^v$​dv =${\int }_0^{\mathrm{t}}\frac{\mathrm{Fdt}}{{\mathrm{m}}_0-\mathrm{\mu t}}$

∴v = $\frac{\mathrm{F}}{-\mathrm{\mu }}{{\left[\ln ({\mathrm{m}}_0-\mathrm{\mu t})\right]}_0}^{\mathrm{t}}$
or v = $\frac{\mathrm{F}}{\mathrm{\mu }}\ln \left(\frac{{\mathrm{m}}_0}{{\mathrm{m}}_0-\mathrm{\mu t}}\right)$
From eq. (i), acceleration of the cart 
a = dv/dt ​= F/m
or a = $\frac{\mathrm{F}}{{\mathrm{m}}_0-\mathrm{\mu t}}$