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We know that work done ON the spring is given by the integral $$\int kx \, dx =\frac12kx^2$$ if we start from $x=0$ .But what if I apply more force on the spring.Wouldn't I be doing more work on it? Why isn't this integral encapsulating this idea? Why should I assume that I just applied a little more force than the spring to move it? Any help will be seriously appreciated.

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The integral is derived from the assumption that external force that is working is equal in magnitude to the equilibrium elastic force of the spring $-kx$. The resulting formula $\frac{1}{2}k(x_2^2-x_1^2)$ is giving the actual work done correctly only for a special case where the spring is moving so slowly during the change that its kinetic energy is negligible. If the spring's mass is small and the end of the spring is moving only with a small speed, this is indeed the case.

Of course it is possible that the work done by the external force is greater than that. This happens if the force pulling the spring is greater in magnitude than the value of equilibrium elastic force $-kx$. This will happen when the end of the spring is moving very fast, so the rest of the spring is not experiencing the same pulling force. The surplus work will be stored as kinetic energy of the spring and will manifest as mutual motion of parts of the spring.

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  • $\begingroup$ So even if I apply a force larger than the force exerted by the spring,the work done on the spring to displace it is still given by that integral? $\endgroup$ – Shannu Shannu Jan 10 at 15:51
  • $\begingroup$ One cannot apply force larger than the force exerted by the spring, because the two are always the same magnitude (due to Newton's 3rd law). However, one can exert force greater than what Hooke's law would predict ($kx$). Then net work done on the spring is greater than integral of $kx$. $\endgroup$ – Ján Lalinský Jan 10 at 16:43
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Work done on the srping captures everything, but you're missing things.

1) Work done is the integral $\int \vec{F}\cdot d\vec{x}$. The integral can be regarded either way as indefinite integral (plus a constant, don't forget the constant), or a definite integral between the final and initial points.

I'll work in 1 dimension hereinafter.

2) Suppose you do the minimum force. You must cancel out Hooke's law, and you do it creating a force equal in magnitude and opposite to the elastic force:

$$F_{you}=-F_{spring}=+k x$$

The work done by your force on the spring is

$$W_{you}=\int_a^b F_{you}\cdot dx =\int_a^b +kx \ dx = \frac{1}{2} k \cdot (\Delta x)^2 $$

3) But now there are two forces. You can also account for the work done by the spring:

$$W_s=\int_a^b F_{spring}\cdot dx =\int_a^b -kx \ dx = -\frac{1}{2} k \cdot (\Delta x)^2 $$

4) The total work on the spring is

$$W_{total}=W_{you}+W_{spring}=0$$

That means no total work, and that's why there is no gain in kinetic energy at all.

5) If you do more force than $kx$, you can split it in two terms: $F_{you}=F_{against}+F_{extra}=+kx + F_{extra}$

The "against" component cancels out the spring's force. The extra part also does work.

The total work is now $W_{Total}=W_{against}+W_s+W_{extra}$. The spring and the against cancel out, and the remaining extra part is the net work done. This results in an increment of kinetic energy

$$W_{Total}=W_{extra}>0; \qquad W_{Total}=\Delta E_k$$

6) The elastic force is conservative. That means that you can write $W_s$ as "minus $\Delta E_p$", for some $E_p$. Of course this potential energy turns out to be $E_p=\frac{1}{2}kx^2$, and $W_s=-\Delta E_p$, as it should be, but this is not the only component of work.

Everything makes sense here.

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  • $\begingroup$ If you didn't understand, tell me. $\endgroup$ – FGSUZ Jan 11 at 23:36

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