Identical springs of steel and copper are equ | vedclass

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(B) The work done in stretching a spring is given by $W = \frac{1}{2} k (\Delta l)^2$,where $k$ is the spring constant and $\Delta l$ is the extension

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Copper spring

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(B) The work done in stretching a spring is given by $W = \frac{1}{2} k (\Delta l)^2$,where $k$ is the spring constant and $\Delta l$ is the extension.Since the springs are identical,they have the same dimensions ($l$ and $A$). The spring constant $k$ is given by $k = \frac{YA}{l}$,where $Y$ is Young's modulus.Therefore,$W = \frac{1}{2} \left( \frac{YA}{l} ight) (\Delta l)^2$.Since $A, l,$ and $\Delta l$ are constant for both springs,we have $W \propto Y$.Because the Young's modulus of steel $(Y_{	ext{steel}})$ is greater than that of copper $(Y_{	ext{copper}})$,it follows that $W_{	ext{steel}} > W_{	ext{copper}}$.Wait,let us re-evaluate: If the springs are stretched by the same *force* $F$,then $\Delta l = \frac{Fl}{AY} \propto \frac{1}{Y}$. The work done $W = \frac{1}{2} F \Delta l \propto \Delta l \propto \frac{1}{Y}$. Since $Y_{	ext{steel}} > Y_{	ext{copper}}$,then $W_{	ext{steel}} Thus,more work is done on the copper spring.

Identical springs of steel and copper are equally stretched. On which more work will have to be done?

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