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.
Related questions
A crate dragged across a rough floor loses mechanical energy because friction is:Conservation of mechanical energy is the form the work-energy theorem takes when the forceCalculating work for a conservative force is easier than for a complicated path because itA force acting entirely perpendicular to an object's displacement does how much work?Mechanical energy transferred into or out of a system equals the work done by an external Winding up a toy or an old-fashioned watch stores energy by doing work against its:Which three energy examples are named together?The kinetic energy in the work-energy theorem is specifically the translational kinetic en