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I have been stuck in a conceptual problem about the sign of the work. For example, suppose that we have a mass $m$ on a spring such that the equilibrium position is at $x=0$, and we stretch the spring to the position $x_0$. At the displacement $x$, the force on the mass is $-kx\hat{i}$ and the differential displacement is $\vec{dx}=dx\hat{i}$. So, the work done by the spring on the mass in this first situation is $$W_1=\int_{0}^{x_0}(-kx)\,dx=-\frac{kx_0^2}{2}$$Now, if we release the spring, it will oscillate around the position $x=0$. My doubt is the work in this situation, that is, the work done by the spring on the mass from the position $x_0$ to the position $x=0$. In this case, I believe that the force is still $-kx\hat{i}$ because the spring is still stretched . But, the differential displacement is now $\vec{dx}=-dx\hat{i}$. So, the work done by the spring on the mass from the point $x_0$ to the point $x=0$ is $$W_2=\int_{x_0}^{0}(-kx)\,(-dx)=-\frac{kx_0^2}{2}$$Although I think that's wrong, since I read in my textbook that the work in the second situation should be positive, I have not understood what I am doing wrong yet. Could you help me, please? In addition, sorry if my English may be broken, it's my second language and I am still working on it.

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  1. You are right, the force always opposes displacement.
  2. The order of integration limits $W_2 = \int_{x_0}^0[...]$ takes care of the direction of displacement. No need for an extra sign in front of $dx$. With this, $W_1 = -W_2$ as it should.
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  • $\begingroup$ I think that he was right when used the vector displacement ( "dx with a sign"). The problem is that the total force is in the positive direction in the first situation, and in the negative direction in the second situation $\endgroup$ – Lucas Freitas Mar 13 at 14:25
  • $\begingroup$ @LucasRodrigues There is no need to consider any other external forces if one is interested solely in the work done by the spring / on the spring. Sure, you may take a step back and calculate the work by evaluating a line integral. But when you plug in the definitions you'll see that it's equivalent to the above expressions. $\endgroup$ – Nephente Mar 13 at 15:21

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