Correct Answer - A::C
From conservation of linear momentum `(1+2) v = (6 xx 1)+(2-3)`
`:. v = 4m//s` (of both the blocks)
from work energy theorm
i.e., `W_("total") = Delta KE` on `1 kg` block,
`W_(f) = (1)/(2) xx 1 xx (4^(2) - 6^(2)) = - 10 J`
on `2 kg` block `W_(f) = (1)/(2) xx 2(4^(2) - 3^(2)) = + 7 J`
`:.` Net work done by friction is ` - 3 J`.