Solve the following :


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.


$F_{1} \neq F_{2}$

$a_{m 1}=\frac{F_{1}-F_{2}}{m_{1}+m_{2}} a_{m 2}=\frac{F_{2}-F_{1}}{m_{1}+m_{2}}$

Net force of $m_{1}=F_{1}-m_{1} a_{m 1}$

$F_{1}^{\prime}=\frac{m_{2} F_{1}+m_{1} F_{2}}{m_{1}+m_{2}}$

Net force of $m_{2}=F_{2}-m_{2} a_{m 2}$

Use C.O.E.L,

$F_{1}^{\prime} x_{1}+F_{1}^{\prime} x_{2}=\frac{1}{2} k\left(x_{1}+x_{2}\right)^{2}$

$\left(\frac{m_{1} F_{2}+m_{2} F_{1}}{m_{1}+m_{2}}\right)\left(x_{1}+x_{2}\right)=\frac{1}{2} k\left(x_{1}+x_{2}\right)^{2}$

$x_{1}+x_{2}=\frac{2}{k} \frac{m_{2} F_{1}+m_{1} F_{2}}{m_{1}+m_{2}}$

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