I found this answer in another forum: \begin{align} W&=\int \mathbf F\cdot {\rm d}\mathbf x \\ &=-\int_0^xkx\,{\rm d}x \\ &=\frac{1}{2}kx^2 \end{align}

I don't understand why we have to set the limits $[0,\,x]$. When deriving kinetic energy, for example, it is enough to merely substitute $F$ but here you need to substitute $F$ and set limits, why?

  • $\begingroup$ -1. Unclear. What other limits would you expect to use instead of [0,x]? Integrals require limits. $\endgroup$ Oct 4, 2017 at 16:46
  • $\begingroup$ Do you know how the calculation of work (shown in your question) is related to spring potential energy? Do you know the definition of potential energy? $\endgroup$
    – Bill N
    Oct 4, 2017 at 17:27
  • $\begingroup$ Possibly useful: physics.stackexchange.com/q/327303/25301 $\endgroup$
    – Kyle Kanos
    Oct 5, 2017 at 10:13

1 Answer 1


Basically you calculate the work needed to extend the spring from 0 to $x$. This work corresponds to the (potential) energy stored in the spring due to the extension.

You could as well calculate the work needed to extend from some arbitrary fixed $x_0$ to $x$ and you would get the same result shifted by a constant term $\frac{1}{2}kx_0^2$. Potential energy is only defined up to a constant. When you do calculations all that enters will be derivatives of the potential so constants will disappear.

For convenience you can take $x_0=0$ and you are left with the relevant part, i.e. the $x$-dependence of the potential.

PS: When deriving the kinetic energy you should have the same, because according to the work-energy theorem the work done should be equal to the change in kinetic energy:

$$\int m\mathbf{a}\cdot\mathbf{dr}=m\int\frac{\mathbf{dv}}{\textrm{dt}}\cdot\mathbf{dr}\\ =m\int_{v_0}^{v} \mathbf{v}\cdot\mathbf{dv}\\ =\frac{1}{2}mv^2-\frac{1}{2}mv_0^2$$


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