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A long spring is stretched by $2 \mathrm{~cm}$ and its potential energy is $U$. If the spring is stretched by $10 \mathrm{~cm}$; its potential energy will be
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Verified Answer
The correct answer is:
$25 U$
The potential energy of a stretched spring is
$$
U=\frac{1}{2} k x^{2}
$$
Here, $k=$ spring constant, $x=$ elongation in spring. But given that, the elongation is $2 \mathrm{~cm}$.
So, $U=\frac{1}{2} k(2)^{2}$
$$
\Rightarrow \quad U=\frac{1}{2} k \times 4 ...(i)
$$
If elongation is $10 \mathrm{~cm}$ then potential energy
$$
\begin{aligned}
U^{\prime} &=\frac{1}{2} k(10)^{2} \\
U^{\prime} &=\frac{1}{2} k \times 100 ...(ii)
\end{aligned}
$$
On dividing Eq. (ii) by Eq. (i), we have
$$
\frac{U^{\prime}}{U}=\frac{\frac{1}{2} k \times 100}{\frac{1}{2} k \times 4}
$$
or $\frac{U^{\prime}}{U}=25 \Rightarrow U^{\prime}=25 U$
$$
U=\frac{1}{2} k x^{2}
$$
Here, $k=$ spring constant, $x=$ elongation in spring. But given that, the elongation is $2 \mathrm{~cm}$.
So, $U=\frac{1}{2} k(2)^{2}$
$$
\Rightarrow \quad U=\frac{1}{2} k \times 4 ...(i)
$$
If elongation is $10 \mathrm{~cm}$ then potential energy
$$
\begin{aligned}
U^{\prime} &=\frac{1}{2} k(10)^{2} \\
U^{\prime} &=\frac{1}{2} k \times 100 ...(ii)
\end{aligned}
$$
On dividing Eq. (ii) by Eq. (i), we have
$$
\frac{U^{\prime}}{U}=\frac{\frac{1}{2} k \times 100}{\frac{1}{2} k \times 4}
$$
or $\frac{U^{\prime}}{U}=25 \Rightarrow U^{\prime}=25 U$
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