In this case we apply Newton's second law of motion:

∑ F = m*a

At the junction between the mass m2 and the rope acts the weight of the mass m2 (W =m2g), at this point, the rope is subjected to tension T2, directed opposite to the weight of the mass m2. At the junction between the rope and the mass m1, the rope is subjected to a T1 voltage, which is equal to the tension T2. In these conditions we can write for the second law:

Let's consider as positive, the forces directed in the sense of clockwise and negative for the opposite direction. In these conditions we can write for the second law:

T1 – T2 + (m2*g) = m*a

As T1 = T2, then:

m2*g = m*a

The total mass of the system is m = (m1 + m2), therefore:

m2*g = (m1 + m2)a

a = m2*g/(m1 + m2) = (1.2)(9.8)/(2 + 1.2)

a = 3.67 m/s^2

To find the tension in the rope, we can apply the second law to the mass m1:

T = m1*a = 2(3.67)

T = 7.74 N

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