A cart loaded with sand moves along a horizontal floor due to a constant force $F$ coinciding in direction with the cart’s velocity vector. In the process, the sand spills through a hole in the bottom with a constant rate $\mu kg/s.$ If at the initial moment $t=0$ the cart with loaded sand had mass $m_0$ and its velocity was equal to zero. Friction is to be neglected.

In this problem, the sand spills through a hole at the bottom of the cart. Hence, the relative  velocity of the sand $v_r$ will be zero, because it will acquire the same velocity as that of the cart  at the moment of separation.

$v_{rel}=0$

Apply $F+v_{rel}\frac{dm}{dt}=m\frac{dv}{dt}\Rightarrow F+0\times \left(\frac{dm}{dt}\right)=(m_0-\mu t)\frac{dv}{dt}\dots \dots \dots .. \left(i\right)$

$\Rightarrow (m_0-\mu t)\frac{dv}{dt}=F or{\int }_0^{v}dv={\int }_0^t\frac{Fdt}{m_0-\mu t}$

$\Rightarrow v=\frac{F}{-\mu }{\left[\ln (m_0-\mu t)\right]}_0^{t}orv=\frac{F}{\mu }\ln \left(\frac{m_0}{m_0-\mu t}\right)$

From Eq (i), acceleration of the cart,  $a=\frac{dv}{dt}=\frac{F}{m}ora=\frac{F}{m_0-\mu t}$