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I've done enough research before asking this question. The work done by a spring is defined as $$W_\mathrm{spring}=\left|\frac{1}{2}kx^2\right|$$ Where $ k$ is the spring constant and $x$ is the distance moved by the spring. But when I tried to derive the eqñ I'm getting $|kx^2|$ and no half is present. I took into consideration that there is displacement at both ends of the spring even when the force is exerted at only one end but still I end up at this eqñ. Can anyone help me out?

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  • $\begingroup$ Any work done by a force is defined as $\int_i^f \mathbf F \cdot *d \mathbf x*$ where $i$ and $f$ represent initial and final coordinates. And, for your case, remember Hooke's Law. $\endgroup$ – frosh Dec 28 '15 at 3:42
  • $\begingroup$ "when I tried to derive the eqñ I'm getting $kx^2$" - How did you derive the equation? $\endgroup$ – Max Payne Dec 28 '15 at 5:32
  • $\begingroup$ "I've done enough research..." apparently not. This is a trivial situation, explained in painful detail in any first-year physics text. $\endgroup$ – Carl Witthoft Dec 28 '15 at 13:59
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Say $W$ is:

$$W=\frac{1}{2}kx^2$$

Then:

$$\frac{dW}{dx}=\frac{1}{2} k \times (x^2)'=\frac{1}{2} k \times 2x=kx$$.

But is it was restoring force $F$ you were looking for, then:

$$F=-\frac{dW}{dx}=-kx$$

Inversely:

$$W=\int Fdx=\int(-kx)dx=-k\int xdx= -k\frac{1}{2}x^2=-\frac{1}{2}kx^2$$

(If integrated between the correct boundaries). The sign is a matter of convention.

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    $\begingroup$ He wants to derive the equation you already accepted in advance, maybe reverse of what you did with the usage of integral might helpful. In addition, when one appreciates that integral is the area under a curve it'd be helpful to draw a plot of $F=kx$. $\endgroup$ – frosh Dec 28 '15 at 3:51
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Any work done by a force is defined as $\int_i^f$ $F$ $dx$, where $i$ and $f$ represent initial and final positions. And, for your case, remember Hooke's Law ($F=-kx$). $$ \int_i^f{-kxdx}=-\frac{1}{2}kx_f^2-(-\frac{1}{2}kx_i^2)=\frac{1}{2}k(x_i^2-x_f^2) $$

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