Why is the formula for energy stored in a spring different from other forces?

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The energy stored in a spring is calculated using the formula 0.5*Fd, which represents the area under the force versus displacement graph. This differs from the work done by constant forces, where W=Fd applies, as the area under the curve is rectangular only when the force is constant. In the case of a spring, the restoring force varies with displacement according to Hooke's Law (F = -kx), making the force a function of position. The area under the graph for a spring forms a triangle, necessitating the use of 0.5*Fd to find the work done. Understanding this concept can be enhanced through integral calculus, although basic area calculations can suffice for simpler cases.
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The energy stored in a spring is found via 0.5*Fd, since this is given by the area under the graph. Why, however, is it usually W=Fd, not W=0.5Fd, as in the spring for other forces?


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I assume you are talking about a force versus displacement graph? In which case the work done is the area under the curve.

For the special case of a stretched spring, that graph forms a right triangle: the base has length "d" (d=total displacement); the height equals "kd". What's the area of a triangle?
 
W = Fd is valid only if the force is constant. Can you see that ONLY if the force is constant will the area "under" the graph be given by F*d? That's because this area will be rectangular. If the force varies, then the area will be some other more complicated shape.

In the case of a spring, the restoring force of the spring is NOT constant. It varies depending on how much the spring is compressed. In other words, it varies as a *function* of position. If x is the displacement of the spring from it's equilibrium position, then F = -kx (Hooke's Law). So F is a function of x.

It is possible to derive the formula for the elastic potential energy stored in the spring using integral calculus. If you don't know integral calculus, don't worry about it for now.

Edit: Or in this case, as Doc Al pointed out, the function is simple enough that you can calculate the area without resorting to integral calculus. But what I was trying to get at is that right now you're just calculating the area under the graph because presumably your teacher is telling you that that will give you the amount of work done. Knowledge of integral calculus allows you to understand the most general mathematical definition of work and therefore to understand WHY it is given by the area under the graph.
 
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