Q127 / 160 · AP EAMCET 2023 (18 May Shift 2) · Physics
A spring has a spring constant $200 \mathrm{Nm}^{-1}$. If it is stretched by $1 \mathrm{~cm}$ then the potential energy stored in it is
- A$100 \mathrm{~J}$
- Correct answerB$0.01 \mathrm{~J}$
- C$10 \mathrm{~J}$
- D$1 \mathrm{~J}$
Solution
Spring constant, $\mathrm{k}=200 \mathrm{~N} / \mathrm{m}$ Stretched length, $x=1 \mathrm{~cm}=0.01 \mathrm{~m}$ Potential energy stored is given as:
$$
\begin{aligned}
& \mathrm{V}=\frac{1}{2} \mathrm{kx}^2=\frac{1}{2} \times 200 \times 0.01 \times 0.01 \\
& \mathrm{~V}=0.01 \mathrm{~J}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.