# Wrong calculation of work done on a spring, how is it wrong?

So I would have thought that this would be how you derive the work on a spring: basically the same way you do with gravity and other contexts, use $$W=\vec{F}\cdot \vec{x}.$$ If you displace a spring by $x$, then it exerts a force $-k x$, so $F=-kx$, since the displacement is $x$.

So $$W=-kx^2.\qquad \leftarrow\text{ (however, apparently wrong!)}$$

I've seen the correct derivation of work in a spring (with an extra half) and don't doubt that it's correct, but also don't see where my logic fails in this alternate derivation.

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Gravity does not change with distance (much), but spring force does (a lot). – ja72 Jul 17 '14 at 20:40

You may be imagining that if you push with constant force $F$, the spring will compress until the spring has such a resistive force.

But since the spring was not counteracting that force, your constant force $F$ was accelerating the mass. Upon reaching the point where the spring has force $F$ as well, the mass does not stop but has a speed such that $KE = \frac{1} {2} k x^2$. So the work your hand does (when pushing with constant force) is correct, but half of the work is put into the spring potential energy and the other half is put in kinetic energy of the mass.

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That was an EXTREMELY insightful answer, thank you! – Addem Jul 18 '14 at 0:24

Since the force is a function of distance, you need to integrate:

$$F = kx\\ W = \int F\ dx\\ W = \int k\ x\ dx\\ W = \frac12kx^2$$

Your work considered the force to be constant - and that's not how springs work.

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Ok, it kind of makes sense that I wasn't considering a constant force. I can certainly see that from the spring's perspective: at a displacement $x$, I'm exerting $-kx$, so if you keep doing this out to a point $x_{0}$ then the work is the integral under the curve. – Addem Jul 17 '14 at 20:46
But I don't quite get it from the human hand's perspective: I push with a constant force of F throughout the motion of the spring until it rests where the spring force cancels my hand's force. From that perspective, shouldn't W=F*d = kx^2? I think this is a general misunderstanding that I have about work, even in the context of just pushing an object on a frictionless surface. – Addem Jul 17 '14 at 20:53
No - you don't push with constant force F against the spring. The spring becomes stiffer as you push it deeper. If you think you are experiencing a constant force, it might be because of inertia of the spring. When you push an mass on a frictionless surface with a constant force, it will accelerate and the work you do becomes kinetic energy. But here we assume "slow compression", ignore kinetic energy, and the force is just proportional to displacement. I hope that clears up the misunderstanding - at least a little bit... – Floris Jul 17 '14 at 21:16
@qwartz - note that I left out vector notation, and that I explicitly said "add signs as needed" since you can argue (I think) that the work done is positive, and that the potential energy stored is positive, if you choose your conventions that way. It depends on whether you think "work done by" or "work done on", etc. - I find I get on fine without worrying about the signs, as I add them based on the physics of the situation (not the math). I find it is more reliable, but that's a personal choice, and it doesn't affect the understanding of this particular question / answer, I believe. – Floris Jul 17 '14 at 22:29

The factor $\frac{1}{2}$ is due to the integral.

The wrong sign of yours is due to the fact that you have to counter the force of the spring. So the Force if the Spring is $-kx$, but you have to pull in the direction it is extended, so apply the force $kx$, therefore the energy is positive $W=\frac 1 2 k^2 x$

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in w=F.X, x is displacement of center of mass of the body not the displacement of system. here also force is 'kx' and displacement of c.o.m. is 0.5*x,so work done will be 0.5kx^2.

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