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Is the work done by external constant force = to potential energy stored in the spring.

i.e. $Fx=\tfrac{1}{2}Kx^2$

But we know $F=kx$ but then $Fx = Kx^2$

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    $\begingroup$ Be careful about merely multiplying both sides of Hooke's Law by "x" to arrive at spring potential energy. It's just coincidence that you arrive at the correct equation by doing this, and any physics teacher worth his salt would deduct test points for such a derivation because it indicates a lack of understanding of the underlying principles involved. $\endgroup$ – David White Mar 12 '19 at 19:42
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A differential amount of work is

$$dw=F.dx$$

If $F$ and $x$ are in the same direction as in the case of the spring, then the amount of work done in compressing the spring is

$$W=\int F.dx=\int (kx)dx=\frac {kx^2}{2}$$

Hope this helps.

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Well see the definition of change in potential energy of the system.It is defined as the negative of work done by internal conservative forces of the system.

Here considering internal forces of the system we have only the spring force which varies with displacement from the reference position.

Now you have applied the equation for work done as F.X which is wrong to apply here as the force is variable in the system.

Now the infinitesimal work done by the spring force in extending the spring is ( -kx.dx ).When you will integrate this equation from 0 to x then you would get the required relation.

Hope this helps.

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  • $\begingroup$ I'm still confused... Isn't the work done by external agent equal to Fx even though the force is constant? $\endgroup$ – Satyam Dwivedi Mar 22 '19 at 15:06
  • $\begingroup$ See a very well description of potential energy and work done is given in CONCEPTS OF PHYSICS by HC VERMA volume 1. $\endgroup$ – user213933 Mar 22 '19 at 15:08
  • $\begingroup$ Here spring force is variable as it varies with displacement from the equilibrium position so work done is given as integration of kx.dx . $\endgroup$ – user213933 Mar 22 '19 at 15:10

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