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

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SUMMARY

The discussion centers on calculating the extra work required 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 these methods incorrect. The correct approach involves using the work-energy principle W = 1/2 kx^2 to determine the spring constant (k) and subsequently calculate the additional work needed. The user ultimately resolved the issue independently.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with work-energy principles in physics
  • Knowledge of the formula W = 1/2 kx^2
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the derivation and application of Hooke's Law in various contexts
  • Learn how to calculate spring constants using experimental data
  • Explore the concept of potential energy stored in springs
  • Investigate real-world applications of spring mechanics in engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of Hooke's Law applications.

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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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