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A system consists of two springs connected in series and each having the spring constant $10 \mathrm{Nm}^{-1}$. The minimum work required to stretch this system by $\mathrm{l} \mathrm{cm}$ in erg is
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$2500$
For series combination, effective force constant is given by
$\frac{1}{K_{\text {eff }}}=\frac{1}{K_1}+\frac{1}{K_2}$
Here, $\quad K_1=K_2=10 \mathrm{~N} / \mathrm{m}$
$\Rightarrow \quad \frac{1}{K_{\text {eff }}}=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}$
or $\quad K_{\text {eff }}=5 \mathrm{~N} / \mathrm{m}=\frac{5 \times 10^5}{100} \mathrm{dyne} / \mathrm{cm}$
$\Rightarrow \quad K_{\text {eff }}=5 \times 10^3$ dyne $/ \mathrm{cm}$
Work done in stretching this spring,
$W=\frac{1}{2} K_{\mathrm{eff}} x^2=\frac{1}{2} \times 5 \times 10^3 \times 1^2=2500 \mathrm{erg}$
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