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The elastic potential energy stored in a spring of spring constant $k$ when stretched by displacement $x$ is:

A$kx$
B$\dfrac{1}{2} k x^2$
C$k x^2$
D$\dfrac{1}{2} k^2 x$
Answer & Solution
Correct answer: B. $\dfrac{1}{2} k x^2$
Force needed to stretch a spring by $x$ is $F = kx$ (Hooke's law). Work done in stretching from 0 to $x$: $U = \int_0^x kx' \, dx' = \dfrac{1}{2} k x^2$. The $\tfrac{1}{2}$ shows up because force grows linearly during the stretch — average force is $kx/2$, total work is $kx/2 \cdot x = \tfrac{1}{2} k x^2$. Option A ($kx$) is the *force*, not the energy.
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