Sum

In the following figure, m_{1} = 5 kg, m_{2} = 2 kg and F = 1 N. Find the acceleration of either block. Describe the motion of m_{1} 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,

m_{1}a = m_{1}g + F − T ...(i)

Again, from the free-body diagram,

m_{2}a = T − m_{2}g − 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/s

After the string breaks, m

m

5a = 1 + 5g

\[\Rightarrow a = \frac{5g + 1}{5}\]

\[ = g + 0 . 2 m/ s^2\]

^{2}.After the string breaks, m

_{1}moves downward with force F acting downward. Then,m

_{1}a = F + m_{1}g5a = 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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