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A long string is stretched by $2 \mathrm{~cm}$ and the potential energy is $\mathrm{V}$. If the spring is stretched by $10 \mathrm{~cm}$, its potential energy will be
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The correct answer is:
$25 \mathrm{~V}$
$\quad V=\frac{1}{2} k(x)^{2}=\frac{1}{2} k(2)^{2}$ or $k=\frac{2 V}{4}=\frac{V}{2}$
$$
\mathrm{V}^{\prime}=\frac{1}{2} \mathrm{k}(10)^{2}=\frac{1}{2} \times\left(\frac{\mathrm{V}}{2}\right)(10)^{2}=25 \mathrm{~V}
$$
$$
\mathrm{V}^{\prime}=\frac{1}{2} \mathrm{k}(10)^{2}=\frac{1}{2} \times\left(\frac{\mathrm{V}}{2}\right)(10)^{2}=25 \mathrm{~V}
$$
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