Work & Hooke's Law Homework: Find Speed of Mass at Equilibrium Position

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SUMMARY

The discussion centers on calculating the speed of a mass at the equilibrium position of a vertical spring with a spring constant of k=1000 N/m and a mass of M=5.0 kg. The spring is compressed by 25 cm before release. The total energy gain calculated is 19 J, leading to a kinetic energy of 6.75 J and a speed of 1.64 m/s. However, the correct speed at equilibrium, as per the answer booklet, is 2.76 m/s, indicating a miscalculation in energy transfer or potential energy considerations.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Knowledge of energy conservation principles in mechanical systems
  • Familiarity with kinetic and potential energy equations
  • Basic algebra for solving equations involving energy
NEXT STEPS
  • Review the derivation of Hooke's Law and its applications in spring mechanics
  • Study energy conservation in mechanical systems, focusing on potential and kinetic energy
  • Learn how to correctly apply the work-energy principle in spring systems
  • Practice problems involving energy transformations in vertical spring scenarios
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of spring dynamics and energy calculations.

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Homework Statement


A vertical spring of negligible mass and spring constant k=1000N/m has a small object of mass M=5.0kg placed on its top. The spring is held compressed by distance of 25 cm from equilibrium position. The spring is released. Find speed of mass at equilibrium position.


Homework Equations



W spring=1/2kdeltax^2
W gravity=deltax*5*9.8
Total energy gain of mass=19J
Energy transferred into potential: 12.25J=mgh
Energy transferred into kinetic: 6.75J ---->v=1.64m/s


The Attempt at a Solution



The answer booklet says the answer is 2.76m/s!
Where did I go wrong?
 
Physics news on Phys.org
I don't know. The 19J seems too low. I got about 32 J. You didn't subtract the potential energy change twice, did you?

Chet
 
W spring=1/2kdeltax^2
When evaluating this, don't overlook that power of 2.
 

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