A system consists of two springs connected in series and each having the spring constant $10…

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
  1. $1500$
  2. $2000$
  3. $3000$
  4. $2500$

Solution


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, $W=\frac{1}{2} K_{\mathrm{eff}} x^2=\frac{1}{2} \times 5 \times 10^3 \times 1^2=2500 \mathrm{erg}$

Asked in: AP EAMCET 2022 (08 Jul Shift 2)

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