# Work and Area under a Curve relating to Hooke's Law

If it takes work W to stretch a Hooke’s-law spring (F = kx) a distance d from its unstressed length, determine the extra work required to stretch it an additional distance d (Hint: draw a graph and give answer in terms of W!).

I don't understand why the answer is not 2W since Force is proportional to x, or even how to begin using a graph to disprove why the answer is not 2W. Any help would be greatly appreciated.

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Even without a graph, the answer is straightforward. The potential energy stored in a spring is proportional to the square of the difference b/w stretched and unstretched length,

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

Thus, the work required to stretch it (mind you, you have to do this slowly so that the Kinetic Energy isn't changed) would be:

$$W = \frac{1}{2}k(2d)^2 - \frac{1}{2}k(d)^2 = \frac{3}{2}kd^2 = 3W_0$$

This is just straightforward conservation of energy.

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Ah, how did I miss something so easy? Thanks! – Joe Feb 16 '12 at 1:49
@Joe You might also want to see this question – pewfly Feb 16 '12 at 2:13