Calculating Work Done by a Spring: A Simple Solution

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SUMMARY

The discussion focuses on calculating the work done by a spring with a force constant of 2.5 x 10^3 N/m when stretched. The work done in stretching the spring by 6.0 cm is calculated to be 4.5 J, while the additional work for stretching it by another 2.0 cm is 3.5 J. The correct formula used is W = 1/2 k x^2, where k is the spring constant and x is the displacement in meters. A common mistake noted was the conversion of centimeters to meters, which led to incorrect calculations.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with the work-energy principle
  • Basic knowledge of unit conversions, specifically from centimeters to meters
  • Proficiency in using the formula W = 1/2 k x^2
NEXT STEPS
  • Review the concept of Hooke's Law and its applications in physics
  • Practice unit conversions, particularly between centimeters and meters
  • Explore examples of work done by springs in different scenarios
  • Learn about potential energy stored in springs and its implications
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of spring dynamics and work calculations.

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[SOLVED] Easy Spring Constant Question

A particular spring has a force constant of 2.5 x 10^3 N/m. (a) How much work is done in stretching the relaxed spring by 6.0 cm? (b) How much more work is done in stretching the spring an additional 2.0 cm?



I am using W=1/2kx^2 but I am not completely sure.



Well the answer are a. 4.5 J and b. 3.5 J. When I plug in I get W=1/2(2500)(8)^2. I am getting the answers 45000 and 35000. I think I am using the right equation I am just multiplying wrong or something..
 
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Ah man! I can't believe I missed that. Thank you!
 

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