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Work Energy Theorem Application

Two springs ...

Question

Two springs $A$ and $B$ ( $k_{A} = 2 k_{B}$ ) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in $A$ is $E$ , that in $B$ is

E / 2

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2 E

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E

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E / 4

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Solution

The correct option is A $E / 2$
Let $F_{A}$ and $F_{B}$ be the spring forces, $x_{A}$ and $x_{B}$ be the spring deformations, and $E_{A}$ and $E_{B}$ be the elastic potential energies of springs $A$ and $B$ respectively.

Since the stretching forces are of equal magnitudes, so will the spring forces. Thus, we get:
$F_{A} = F_{B}$
$\Rightarrow k_{A} x_{A} = k_{B} x_{B}$
$\Rightarrow 2 k_{B} x_{A} = k_{B} x_{B}$
$\Rightarrow \frac{x_{A}}{x_{B}} = \frac{1}{2}$

Since for a spring, $E = \frac{1}{2} k x^{2}$ , we get:
$\frac{E_{A}}{E_{B}} = \frac{\frac{1}{2} k_{A} x_{A}^{2}}{\frac{1}{2} k_{B} x_{B}^{2}}$

$\Rightarrow \frac{E_{A}}{E_{B}} = \frac{k_{A}}{k_{B}} (\frac{x_{A}}{x_{B}})^{2}$

$\Rightarrow \frac{E_{A}}{E_{B}} = 2 (\frac{x_{A}}{x_{B}})^{2}$

$\Rightarrow \frac{E_{A}}{E_{B}} = 2 (\frac{1}{2})^{2}$

$\Rightarrow \frac{E_{A}}{E_{B}} = \frac{1}{2}$

$∴ E_{B} = 2 E$

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