Question

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. 

Answer 1

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,
\[x_2 = \frac{2F m_1}{k( m_1 + m_2 )}\]