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(D) For springs connected in series,the effective spring constant $K_{	ext{eff}}$ is given by:$\frac{1}{K_{	ext{eff}}} = \frac{1}{K_1} + \frac{1}{K_

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$2500$

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(D) For springs connected in series,the effective spring constant $K_{	ext{eff}}$ is given by:$\frac{1}{K_{	ext{eff}}} = \frac{1}{K_1} + \frac{1}{K_2}$Given $K_1 = K_2 = 10 	ext{ N/m}$,we have:$\frac{1}{K_{	ext{eff}}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}$Thus,$K_{	ext{eff}} = 5 	ext{ N/m}$.To convert $K_{	ext{eff}}$ to $	ext{dyne/cm}$:$1 	ext{ N} = 10^5 	ext{ dyne}$ and $1 	ext{ m} = 100 	ext{ cm}$.$K_{	ext{eff}} = 5 	imes \frac{10^5 	ext{ dyne}}{100 	ext{ cm}} = 5 	imes 10^3 	ext{ dyne/cm}$.The work done $W$ to stretch the system by $x = 1 	ext{ cm}$ is:$W = \frac{1}{2} K_{	ext{eff}} x^2 = \frac{1}{2} 	imes (5 	imes 10^3) 	imes (1)^2 = 2500 	ext{ erg}$.

$A$ system consists of two springs connected in series,each having a spring constant of $10 \text{ N m}^{-1}$. The minimum work required to stretch this system by $1 \text{ cm}$ in $\text{erg}$ is

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