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Step-by-Step Solution
Step 1: Recall the expression for energy stored in a spring
The potential energy $U$ stored in a stretched (or compressed) spring by a force $F$ is given by:
$$
U = \frac{1}{2}\,K\,x^2
$$
Here, $K$ is the spring constant and $x$ is the extension (or compression).
Step 2: Express $x$ in terms of the force $F$ and the spring constant $K$
For a spring, the applied force $F$ relates to its extension $x$ by Hooke's law:
$$
F = K \, x
$$
Therefore,
$$
x = \frac{F}{K}
$$
Step 3: Rewrite the energy in terms of $F$ and $K$
Substituting $x = \frac{F}{K}$ into the energy formula, we get:
$$
U = \frac{1}{2} \, K \left(\frac{F}{K}\right)^2
= \frac{1}{2} \frac{F^2}{K}
$$
Step 4: Relate the spring constants for Springs A and B
Given that $K_A = 2\,K_B$. If the same magnitude of force $F$ is applied to both springs, then the energy stored in each spring is:
$$
E_A = \frac{1}{2}\,\frac{F^2}{K_A}
\quad\text{and}\quad
E_B = \frac{1}{2}\,\frac{F^2}{K_B}
$$
Substituting $K_A = 2\,K_B$ gives:
$$
E_A = \frac{1}{2}\,\frac{F^2}{2\,K_B} = \frac{1}{4}\,\frac{F^2}{K_B}
\quad\text{and}\quad
E_B = \frac{1}{2}\,\frac{F^2}{K_B}
$$
Step 5: Compare the energies $E_A$ and $E_B$
From the above expressions:
$$
\frac{E_A}{E_B} = \frac{\frac{1}{4}\,\frac{F^2}{K_B}}{\frac{1}{2}\,\frac{F^2}{K_B}}
= \frac{1/4}{1/2} = \frac{1}{2}
$$
Thus,
$$
E_B = 2\,E_A
$$
Final Answer
The energy stored in spring B is $2\,E_A$.
