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प्रश्न
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.
उत्तर
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 = \[\left( \frac{1}{2} \right)K( x_1 + x_2 )^2 . . . (2)\]
\[\text{ Equating the equations (1) and (2), we get: } \]
\[ \Rightarrow ( x_1 + x_2 ) = \left( \frac{2F}{k} \right) \ldots(3)\]
As the net external force on the system is zero, the centre of mass does not shift.
\[\therefore m_1 x_1 = m_2 x_2 . . . (4)\]
\[\text{ Using the equations }\left( 3 \right) \text{ and } \left( 4 \right), \text{ we get}: \]
\[ \frac{m_1}{m_2} x_1 + x_1 = \frac{2F}{k}\]
\[ \Rightarrow x_1 \left( 1 + \frac{m_1}{m_2} \right) = \frac{2F}{k}\]
\[ \Rightarrow x_1 = \frac{2F m_2}{k( m_1 + m_2 )}\]
Similarly,
\[x_2 = \frac{2F m_1}{k( m_1 + m_2 )}\]
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