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(B) The potential energy $U$ stored in a spring stretched by a distance $x$ is given by $U = \frac{1}{2} k x^2$, where $k$ is the spring constant.In t

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$4$

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(B) The potential energy $U$ stored in a spring stretched by a distance $x$ is given by $U = \frac{1}{2} k x^2$, where $k$ is the spring constant.In the first case, the extension is $x_1 = 10 \ cm = 0.1 \ m$. Thus, $E = \frac{1}{2} k (0.1)^2 = 0.005 k$.In the second case, the spring is stretched by $10 \ cm$ more, so the total extension is $x_2 = 10 \ cm + 10 \ cm = 20 \ cm = 0.2 \ m$.The new potential energy $E'$ is $E' = \frac{1}{2} k (0.2)^2 = \frac{1}{2} k (4 	imes 0.01) = 4 	imes (\frac{1}{2} k (0.1)^2)$.Therefore, $E' = 4 E$.

When a spring is stretched by $10 \ cm$, the potential energy stored is $E$. When the spring is stretched by $10 \ cm$ more, the potential energy stored in the spring becomes (in $E$)

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