Two springs A and B , having spring constants related by k_A = 2k_B...

The energy stored in a spring stretched by a force $F$ is given by $U = \dfrac{1}{2}kx^2$.

Since $F = kx$, the extension is $x = \dfrac{F}{k}$.

Substituting $x$ in the energy expression, we get $U = \dfrac{F^2}{2k}$.

Given that both springs are stretched by forces of equal magnitudes, $F_A = F_B = F$.

For spring $A$, the stored energy is $E = \dfrac{F^2}{2k_A}$.

For spring $B$, the stored energy is $E_B = \dfrac{F^2}{2k_B}$.

Given $k_A = 2k_B$, we can write $k_B = \dfrac{k_A}{2}$.

Substituting this into the equation for $E_B$ gives:

$E_B = \dfrac{F^2}{2 \left( \dfrac{k_A}{2} \right)} = 2 \left( \dfrac{F^2}{2k_A} \right) = 2E$.