Initially spring is in natural length and bot | vedclass

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Question

Using the work energy theorem,

$\left(F-m_{1} a\right) x_{1}+\left(m_{2} a\right) x_{2}-\frac{1}{2} k\left(x_{1}+x_{2}\right)^{2}=0$

$m_{2}\left

Vedclass · Accepted Answer

$\frac{20}{3} \,cm$

Vedclass · Answer

Using the work energy theorem,

$\left(F-m_{1} a\right) x_{1}+\left(m_{2} a\right) x_{2}-\frac{1}{2} k\left(x_{1}+x_{2}\right)^{2}=0$

$m_{2}\left(\frac{F}{m_{1}+m_{2}}\right)\left(x_{1}+x_{2}\right)=\frac{k}{2}\left(x_{1}+x_{2}\right)^{2}$

$\left(x_{1}+x_{2}\right)=m_{2}\left(\frac{F}{m_{1}+m_{2}}\right)\left(\frac{2}{k}\right)$

$=\frac{1(1)}{1.5}\left(\frac{2}{20}\right)$

Further simplify the above,

$\frac{1}{15} m =\frac{100}{15} cm$

$=\frac{20}{3} cm$

Initially spring is in natural length and both blocks are in rest condition. Then determine Maximum extension is spring. $k=20 N / M$

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