How Much Work Is Needed to Stretch a Spring from 12 cm to 16 cm?

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SUMMARY

The discussion focuses on calculating the work required to stretch a spring from 12 cm to 16 cm, given a relaxed length of 7 cm and a stiffness of 50 N/m. The correct approach involves using the formula for potential energy in springs, specifically ΔPE = 1/2 * k * s², where 's' represents the stretch. The correct calculation shows that the work done is 0.14 J, highlighting the importance of using the correct stretch values rather than the lengths directly.

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  • Understanding of Hooke's Law and spring constants
  • Familiarity with potential energy equations in physics
  • Knowledge of unit conversions, specifically from centimeters to meters
  • Basic algebra skills for manipulating equations
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Homework Statement


A spring has a relaxed length of 7 cm and a stiffness of 50 N/m. How much work must you do to change its length from 12 cm to 16 cm?


Homework Equations


1/2*kss2


The Attempt at a Solution


1/2*50*(.16-.12)2 = 0.04

It shows the solution to be 0.14. It says the change in the energy of the spring is its change in potential energy, so the change in (1/2)kss2. Be sure to use the stretch s, not the length, and use meters, not centimeters.
 
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Looks like an algebra error and a misunderstanding of stretch.
Length 12 is a stretch of 12 - 7 = 5. Length 16 is a stretch of 16 - 7 = 9.
.5k(.09)^2 - .5k(.05)^2 is not equal to .5k(.09-.05)^2.
 

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