The formula for elastic potential energy is $\rm (1/2)kx²$, all that I am asking is why is it not $\rm kx²$. Here is my logic. Let us say that $\rm k$ is $\rm 2\ N/m$ which means that for the spring to move $\rm 1 \ meter$ you have to apply $\rm 2 \ N$ of force, if you want to move the spring $\rm 2 \ meter$, then you have to apply $\rm 2N$ and then $\rm 2 N$ again, so the work should be $\rm 2N×4m=8J$. Where did I go wrong?
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$\begingroup$ Try a site search: there are answers here, here, here, here... $\endgroup$– ChemomechanicsCommented Aug 19, 2021 at 17:24
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$\begingroup$ ...here, here... $\endgroup$– ChemomechanicsCommented Aug 19, 2021 at 17:24
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2 Answers
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$\rm k$ is $\rm 2\ N/m$ which means that for the spring to move $\rm \ 1 meter$ you have to apply $2 \ N$ of force.
That's a misleading version of what $\rm k$ means, because when you first start stretching the spring almost no force is needed. $\rm k= 2 \ N/m$ means that when the spring has an extension of $\rm 1\ m$ the tension in it is $\rm 2 \ N$, and when the extension is $\rm 2 \ m$ the tension is $\rm 4\ N$ and so on. The tension rises proportionally to the extension. As you stretch the spring the force you must apply rises continuously.
The work done stretching the spring (and therefore the elastic PE) is the area under the graph of force against extension. Read all about it in a textbook!
answered Aug 19, 2021 at 10:09
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What you did wrong is assume that the force is constant.
Imagine that you keep contracting the spring: at the beginning it is very easy but then it gets harder and harder. So the work you do depends on the position $x$ you are at, it is not always $2N$ of force that you applay: the $k$ in the spring force means "if I contract the spring by $x$ the force is $-kx$", it does not mean that for each extra meter you need to apply $2N$!
The force of a spring (that you need to balance) is $F=-kx$ so it varies with the distance. If you compress it by one meter with $k=2N/m$ then $F=-(2N/m)*1m=-2N$. If you compress it by another meter (ending up at $x=2 m$) the force of the spring is $F=-(2N/m)*2m=-4N$. Which means that for the first half you did a work smaller than in the second half as the force was smaller! And if you divide the first meter in 2 sectors of 0.5 m in the first sector you do an even smalller work. And if you keep dividing, you notice that in each sector you do a different (and increasing with $x$) amount of work. The total work is the sum of all these works!
A quick argument would be to use the average force that you apply to compress a spring by a quantity $\Delta x$. At the beginning ($x=0$) you need to apply $0$ force. At the end, you need to apply a force $k\Delta x$, then the mean force is $f_m=(k\Delta x+0)/2$ and the mean work is mean force time displacement i.e. $$W_m=f_m\Delta x = {1\over 2}k\Delta x^2$$
A more general derivation is using integrals. The definition of work is, for a displacement $dx$ at a position $x$ the small work you need is $$dW = F(x)dx = kx dx$$
which means that for the total work you need to integrate this
$$W=\int^x_0 dW =\int kx dx={1 \over 2} k x^2$$ which is the potential energy you seek.
