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

AI Thread Summary
A vertical spring with a spring constant of 1000 N/m is compressed by 25 cm with a 5.0 kg mass on top. The total energy gain calculated is 19 J, with 12.25 J attributed to potential energy and 6.75 J to kinetic energy, leading to a calculated speed of 1.64 m/s. However, the answer booklet states the speed at the equilibrium position is 2.76 m/s, prompting confusion about the calculations. A participant suggests that the discrepancy may arise from not properly accounting for the potential energy change. The discussion emphasizes the importance of correctly applying the equations, particularly the spring energy formula.
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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?
 
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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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