How Much Extra Work Is Needed to Stretch a Spring Further?

AI Thread Summary
The discussion revolves around calculating the extra work needed to stretch a Hooke's-law spring an additional 5.76 cm after initially stretching it 12.1 cm with 3.23 J of work. The user initially attempted to use proportions and the work formula W = FD but found their answers incorrect. They considered whether they needed the spring constant (k) or the equation W = 1/2kx^2 for the calculation. Ultimately, the user resolved their confusion and found the solution independently. The thread highlights the importance of understanding the relationship between force, displacement, and work in spring mechanics.
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Homework Statement


It takes 3.23 J of work to stretch a Hooke's-law spring 12.1 cm from its unstressed length. How much extra work is required to stretch it an additional 5.76cm. Answer in Joules.

Homework Equations


F= -kx
W= FD


The Attempt at a Solution


At first, I try using proportions to do this problem:

3.23/12.1 = x/5.76
I got that answer but it was wrong.

So I tries finding the force using w=fd. Then I multiplied that by 5.76 and I got the same answer as doing it the proportion method.

What am I doing wrong. Do I need K or do I need to use the equation W = 1/2kx^2? Thanks in advance.
 
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Oh figured it out, nvm.
 

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