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Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
6
Work and Energy
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6-1: Work, Constant Force
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6-2: Work, Varying Force
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6-3: Kinetic Energy; Work-Energy Principle
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6-4 and 6-5: Potential Energy
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6-6 and 6-7: Conservation of Mechanical Energy
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6-8 and 6-9: Law of Conservation of Energy
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6-10: Power
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Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 27
Q

A spring has a spring constant kk of 88.0 N/m. How much must this spring be compressed to store 45.0 J of potential energy?

A
1.01 m1.01 \textrm{ m}
Giancoli 7th Edition, Chapter 6, Problem 27 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The elastic potential energy of a spring equals one-half times the spring constant, K, times the amount that it is stretched or compressed from its equilibrium position and we square that. And we can solve for x by multiplying both sides by 2 and dividing by K and then take the square root of both sides and we get x is square root of 2 times the potential energy stored divided by the spring constant. So that's square root of 2 times 45 joules divided by 88 newtons per meter and that gives 1.01 meters of compression to store 45 joules.

COMMENTS
By sueqrahn on Sun, 1/10/2016 - 4:15 AM

This answer should only have the units of meters, not meters per second

By Mr. Dychko on Sun, 1/10/2016 - 9:34 PM

Hi sueqrahn, I've corrected the typo. Thanks for spotting that.

All the best,
Mr. Dychko

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Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.