Sum
In the following figure, m1 = 5 kg, m2 = 2 kg and F = 1 N. Find the acceleration of either block. Describe the motion of m1 if the string breaks but F continues to act.
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Solution
Let the acceleration of the blocks be a.
The free-body diagrams for both the blocks are shown below:
From the free-body diagram,
m1a = m1g + F − T ...(i)
Again, from the free-body diagram,
m2a = T − m2g − F ...(ii)
Adding equations (i) and (ii), we have:
\[a = g\frac{m_1 - m_2}{m_1 + m_2}\]
\[\Rightarrow a = \frac{3g}{7} = \frac{29 . 4}{7}\]
\[ = 4 . 2 m/ s^2\]
Hence, acceleration of the block is 4.2 m/s2.
After the string breaks, m1 moves downward with force F acting downward. Then,

m1a = F + m1g
5a = 1 + 5g
\[\Rightarrow a = \frac{5g + 1}{5}\]
\[ = g + 0 . 2 m/ s^2\]
After the string breaks, m1 moves downward with force F acting downward. Then,

m1a = F + m1g
5a = 1 + 5g
\[\Rightarrow a = \frac{5g + 1}{5}\]
\[ = g + 0 . 2 m/ s^2\]
Concept: Newton’s Second Law of Motion
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