Answer
3.6 N
Work Step by Step
$N=mg-P=(2.5\,kg\times9.8\,m/s^{2})-10\,N=14.5\,N$
$f_{s_{max}}=\mu_{s}N=0.40\times14.5\,N=5.8\,N$
Applied force is more than $f_{s_{max}}$ and therefore the block starts to move. In this case, the force of kinetic friction is
$f_{k}=\mu_{k}N=0.25\times14.5\,N=3.6\,N$
