Consider the situation of the previous problem. Suppose the block of mass m_{1} is pulled by a constant force F_{1} and the other block is pulled by a constant force F_{2}. Find the maximum elongation that the spring will suffer.

#### Solution

Given:

Force on block of mass, m_{1} = F_{1 }

Force on block of mass, m_{2} = F_{2}

Let the acceleration produced in mass m_{1} be a_{1}.

\[a_1 = \frac{F_1 - F_2}{m_1 + m_2}\]

Let the acceleration of mass m_{2}_{ }be a_{2}.

\[a_2 = \frac{F_2 - F_1}{m_1 + m_2}\]

Due to the force F_{2}, the mass m_{1} experiences a pseudo force.

\[\therefore \text{ Net force on m}_1 = F_1 + m_1 a_2 \]

\[F' = F_1 + m_1 \times \frac{( F_2 - F_1 )}{m_1 + m_2}\]

\[ = \frac{m_1 F_1 + m_2 F_1 + m_1 F_2 - m_1 F_1}{m_1 + m_2}\]

\[ = \frac{m_2 F_1 + m_1 F_2}{m_1 + m_2}\]

Similarly, mass m_{2} experiences a pseudo force due to force F_{1}.

\[\therefore \text{ Net force on m}_2 = F_2 + m_2 a_1\]

\[F " = F_2 + m_2 \times \frac{( F_1 - F_2 )}{m_1 + m_2}\]

\[ = \frac{m_1 F_2 + m_2 F_2 + m_2 F_1 - m_2 F_2}{m_1 + m_2}\]

\[ = \frac{m_1 F_2 + m_2 F_1}{m_1 + m_2}\]

Let m_{1} be displaced by a distance x_{1} and m_{2} be displaced by a distance x_{2}.

Therefore, the maximum elongation of the spring = x_{1} + x_{2}

Work done by the blocks = Energy stored in the spring

\[\Rightarrow \frac{m_2 F_1 + m_1 F_2}{m_1 + m_2} \times x_1 \times \frac{m_2 F_1 + m_1 F_2}{m_1 + m_2} \times x_2 = \left( \frac{1}{2} \right)k( x_1 + x_2 )^2 \]

\[ \Rightarrow x_1 + x_2 = \frac{2}{k}\left( \frac{m_1 F_2 + m_2 F_1}{m_1 + m_2} \right)\]