Question

Consider the situation of the previous problem. Suppose the block of mass m₁ is pulled by a constant force F₁ and the other block is pulled by a constant force F₂. Find the maximum elongation that the spring will suffer.  

Answer 1

Given:
Force on block of mass, m₁ = F₁ 
Force on block of mass, m₂ =​ F₂

Let the acceleration produced in mass m₁ be a₁.
\[a_1 = \frac{F_1 - F_2}{m_1 + m_2}\]
Let the acceleration of mass m₂ be a₂.
\[a_2 = \frac{F_2 - F_1}{m_1 + m_2}\]
Due to the force F₂, the mass m₁ 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 m₂ experiences a pseudo force due to force F₁. 

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

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

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

\[ = \frac{m_1 F_2 + m_2 F_1}{m_1 + m_2}\]
Let m₁ be displaced by a distance x₁ and m₂ be displaced by a distance ​x₂.

Therefore, the maximum elongation of the spring = x₁ + x₂

Work done by the blocks = Energy stored in the spring

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

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