Consider the situation of the previous problem. Suppose each of the blocks is pulled by a constant force F instead of any impulse. Find the maximum elongation that the spring will suffer and the distance moved by the two blocks in the process.

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Solution

It is given that the force on both the blocks is F.

Let x₁ and x₂ be the extensions of blocks m₁ and m₂ respectively.

Total work done by the forces on the blocks = Fx₁ + Fx₂ ...(1)
∴ Increase in the potential energy of spring = $(\frac{1}{2}) K (x_{1} + x_{2})^{2} . . . (2)$
$Equating the equations (1) and (2), we get : \Rightarrow (x_{1} + x_{2}) = (\frac{2 F}{k}) \dots (3)$

As the net external force on the system is zero, the centre of mass does not shift.
$∴ m_{1} x_{1} = m_{2} x_{2} . . . (4) Using the equations (3) and (4), we get : \frac{m_{1}}{m_{2}} x_{1} + x_{1} = \frac{2 F}{k} \Rightarrow x_{1} (1 + \frac{m_{1}}{m_{2}}) = \frac{2 F}{k} \Rightarrow x_{1} = \frac{2 F m_{2}}{k (m_{1} + m_{2})}$

Similarly,
$x_{2} = \frac{2 F m_{1}}{k (m_{1} + m_{2})}$

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Figure

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