Two bodies of masses 4 kg and 6 kg connected by means of a light string are lying on a smooth horizontal surface. A horizontal pulling force is applied on the lighter body. Two seconds later the string connecting the two masses is cut. After two more seconds if the velocity of the heavier mass is 2 ms^–1, the force initially applied is

When the string is intact, both bodies move as a single body and 
the acceleration a = F/m, where m = 4kg + 6kg = 10kg.
The speed attained in 2 secs is 
v = u + at
 = (F/10)(2 secs)

After 2 secs when the string is cut, the 6kg mass will have no force acting on it. So, it will stay in motion with a uniform velocity which is attained when the string is intact.

Given the velocity of 6kg mass as 2m/s.

$$\begin{gathered} \Rightarrow \mathrm{v} = \mathrm{F}/5 = 2{\mathrm{ms}}^{-1} \\ \therefore \mathrm{F} = 10\mathrm{N} \end{gathered}$$