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
(a)$\Delta \text { P.E. }=\frac{1}{2} k\left(x_2^2-x_1^2\right)$
$=\frac{1}{2} \times 10\left[(0.25)^2-(0.20)^2\right] $
$=5 \times0.45\times0.05=0.1\mathrm{~J}$

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