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Initially spring is in natural length and both the blocks are in rest condition. Then determine the maximum extension in the spring. $K=20 \mathrm{~N} \mathrm{~m}^{-1}$
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Verified Answer
The correct answer is:
$\frac{20}{3} \mathrm{~cm}$
$a=\frac{F}{m_1+m_2}$
By work - energy theorem
$\begin{aligned} & \left(F-m_1 a\right) x_1+\left(m_2 a\right) x_2-\frac{1}{2} K\left(x_1+x_2\right)^2=0 \\ & m_2 \times \frac{F}{m_1+m_2}\left(x_1+x_2\right)=\frac{K}{2}\left(x_1+x_2\right)^2 \\ & \left(x_1+x_2\right)=m_2 \times \frac{F}{m_1+m_2} \times \frac{2}{k} \\ & =\frac{1 \times 1}{1.5} \times \frac{2}{20}=\frac{1}{15} m=\frac{100}{15} \mathrm{~cm}=\frac{20}{3} \mathrm{~cm}\end{aligned}$
By work - energy theorem
$\begin{aligned} & \left(F-m_1 a\right) x_1+\left(m_2 a\right) x_2-\frac{1}{2} K\left(x_1+x_2\right)^2=0 \\ & m_2 \times \frac{F}{m_1+m_2}\left(x_1+x_2\right)=\frac{K}{2}\left(x_1+x_2\right)^2 \\ & \left(x_1+x_2\right)=m_2 \times \frac{F}{m_1+m_2} \times \frac{2}{k} \\ & =\frac{1 \times 1}{1.5} \times \frac{2}{20}=\frac{1}{15} m=\frac{100}{15} \mathrm{~cm}=\frac{20}{3} \mathrm{~cm}\end{aligned}$
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