The force due to m1m_{1}m1 on m2m_{2}m2 is 6N6 N6N
the force due to m2m_{2}m2 on m1m_{1}m1 is 4N4 N4N
To get the acceleration, aaa of the system, we find the net force, FFF .
F=6N−4N=2NF = 6N - 4N = 2NF=6N−4N=2N
total mass =3+4=7= 3 + 4 = 7=3+4=7
F=maF = maF=ma
a=Fm=27=0.286ms−2a = \frac{F}{m} = \frac{2}{7} = 0.286 m s^{-2}a=mF=72=0.286ms−2
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