A cart has mass 2 metric tone and sand of 1 metric tone is inside...

Mass of the cart with sand changes then its acceleration also changes. $a=\frac{F}{m}=\frac{F}{m_0-\mu t}$ where $m$ is mass of system at time $t$, $$\begin{aligned} & m=m_0-\mu t \\ & a=\frac{F}{m_0-\mu t} \quad \text{\text{ or } } \quad \frac{d v}{d t}=\frac{F}{m_0-\mu t}\end{aligned}$$$$\begin{aligned} & \int_0^v d v=F \int_0^t \frac{d t}{m_0-\mu t} \\ & v=\frac{F}{\mu} \ln \left(\frac{m_0}{m_0-\mu t}\right)\end{aligned}$$$Here$F=10 \mathrm{~N} ; \mu=0.5 \mathrm{~kg} / \mathrm{s}m_0=2$metric tone$m_0-\mu t=1$ metric tone \therefore \quad v=\frac{10}{0.5} \ln (2)=20 \times 0.693=13.86 \mathrm{~m} / \mathrm{s}