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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