Calculating Potential Energy Increase in a Massless Spring

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SUMMARY

The potential energy increase in a massless spring when a 28.8 kg mass is attached is calculated using the formula PE = 0.5kx², where k is the spring constant of 4800 N/m. After determining the displacement x as 0.0588 meters, the potential energy is found to be 8.3 Joules. The initial incorrect approach of using 0.5Kx²=mg led to confusion, but the correct method confirms the potential energy increase is indeed 8.3 J.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with potential energy formulas
  • Basic algebra for solving equations
  • Knowledge of gravitational force calculations (mg)
NEXT STEPS
  • Study Hooke's Law and its applications in mechanics
  • Learn about energy conservation principles in spring systems
  • Explore the effects of varying spring constants on potential energy
  • Investigate real-world applications of springs in engineering
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Students in physics, mechanical engineers, and anyone interested in understanding the principles of potential energy in spring systems.

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Homework Statement


A massless spring hangs from the ceiling. How much does its potential energy increase, if a 28.8 kg mass is attached to it? The spring constant is 4800 N/m.




The Attempt at a Solution



I set 0.5Kx^2=mg. I then divided the spring constant by mg. I got 17 J. This really doesn't seem right to me. Any import would be greatly appreciated
 
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-kx = mg

not 0.5kx^2=mg

find x and then proceed
 
Ok so I solved for x and got 0.0588 meters. I then plugged it into the 0.5kx^2 and got 8.3 J. Does this seem correct?
 

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