MCQ
A spring of force constant $10 \mathrm{~N} / \mathrm{m}$ has an initial stretch $0.20 \mathrm{~m}$. In changing the stretch to $0.25 \mathrm{~m}$, the increase in potential energy is about
- ✓$0.1$ joule
- B$0.2$ joule
- C$0.3$ joule
- D$0.5$ joule
✓
Answer
Correct option: A.
$0.1$ joule
$\Delta \text { P.E. }=\frac{1}{2} k\left(x_2^2-x_1^2\right)$
$=\frac{1}{2} \times 10\left[(0.25)^2-(0.20)^2\right] $
$=5 \times0.45\times0.05=0.1\mathrm{~J}$
$=\frac{1}{2} \times 10\left[(0.25)^2-(0.20)^2\right] $
$=5 \times0.45\times0.05=0.1\mathrm{~J}$
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