A chain consisting of 5 links each of mass 1kg is lifted vertically with a constant acceleration of $2m{s}^{-2}$ as shown in figure. The force of interaction between the top link and the link immediately below it will be ______ N. $\left[\mathrm{Take}\mathrm{g}=10{\mathrm{ms}}^{-2}\right]$  

If F is the upward force applied, a is the net upward acceleration and m is the mass of each link. Then, $F-5mg=5maorF=5m(g+a)\mathrm{....}\left(i\right).$ 

If N is force of interaction between the top link and link immediately below it, then 

$\begin{array}{l}ma=F-mg-N\\ N=F-mg-ma=5m(g+a)-m(g+a) \left(U\sin g\right(i\left)\right)\\ N=4m(g+a)=4\times 1\times (10+2)=48N\end{array}$