# A 2.00 kg mass is on a table attached to a string over a pulley which is then attached to another mass (1.20 kg) hanging over the edge. This is much like the set up in lab. There is no friction...

A 2.00 kg mass is on a table attached to a string over a pulley which is then attached to another mass (1.20 kg) hanging over the edge. This is much like the set up in lab. There is no friction between the table and the 2.00 kg mass.

What will be the acceleration of the blocks?

What is the tension in the string?

Can you please include a free body diagram?

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### 1 Answer

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