Additional energy stored in spring

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SUMMARY

The discussion centers on calculating the additional elastic potential energy stored in a spring when a load of 10 N is applied, followed by an extension of 0.040 m. The relevant equations are F = kx and E = 1/2 kx². The correct answer for the additional energy stored is 0.200 J, as confirmed by the answer key, despite initial confusion regarding the calculations leading to 0.600 J. The key takeaway is the importance of accurately applying the formulas for elastic potential energy.

PREREQUISITES
  • Understanding of Hooke's Law (F = kx)
  • Knowledge of elastic potential energy formula (E = 1/2 kx²)
  • Basic concepts of force and extension in springs
  • Ability to perform unit conversions and calculations
NEXT STEPS
  • Review the principles of Hooke's Law in detail
  • Practice problems involving elastic potential energy calculations
  • Explore the relationship between force, extension, and spring constant (k)
  • Investigate real-world applications of springs in engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators looking for examples of spring dynamics and energy calculations.

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Homework Statement


The graph shows the variation with extension x of the load F on a certain spring
upload_2019-1-12_17-3-1.png

A load of 10 N is placed on the spring. How much additional elastic potential energy will be stored in the spring if it is then extended a further 0.040 m?
a. 0.200 J
b. 0.450 J
c. 0.600 J
d. 0.800 J

Homework Equations


F = k x
E = 1/2 k x2

The Attempt at a Solution


Load of 10 N means the extension is 0.04 m and extended further 0.04 m means that the extension is 0.08 m, right?

I got (c) but the answer key is (a) ??

Thanks
 

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songoku said:

Homework Statement


The graph shows the variation with extension x of the load F on a certain spring
View attachment 237172
A load of 10 N is placed on the spring. How much additional elastic potential energy will be stored in the spring if it is then extended a further 0.040 m?
a. 0.200 J
b. 0.450 J
c. 0.600 J
d. 0.800 J

Homework Equations


F = k x
E = 1/2 k x2

The Attempt at a Solution


Load of 10 N means the extension is 0.04 m and extended further 0.04 m means that the extension is 0.08 m, right?

I got (c) but the answer key is (a) ??

Thanks

I think you are correct.
 
Thank you
 

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