Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.