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horizontal component force will provide accleration ...
Fcos30 = 31/2F/2 = FH = force in horizontal direction
ma = 31/2F/2
m=2kg , a = 31/2 so
F = 4N
net force acting in verticle direction = mg+Fsin30 = 22N
normal force bw table & block = 22N
normal force bw ground & table = 22+weight of table
=22+100 = 122N
net force acting on block = [ N2 + (FH)2 ]1/2
N is normal force bw block & table , FH is horizontal comp of applied force ....
= [12 + 222]1/2 = 22.3N
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