Conservation of Energy and Springs. Block dropped onto a spring

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SUMMARY

The discussion revolves around the physics problem of a 2.7 kg block dropped from a height of 4.5 m onto a spring, which compresses by 25.0 cm. The spring constant was calculated to be approximately 4026 N/m, although some participants suggested alternative values around 3000 N/m. The key equations used include potential energy (PE = mgh), kinetic energy (KE = 0.5mv²), and spring potential energy (U(s) = 0.5kx²). The speed of the block at a compression of 15.0 cm was calculated to be 7.52 m/s, prompting further inquiries about the height used in calculations.

PREREQUISITES
  • Understanding of potential energy (PE) and kinetic energy (KE) equations
  • Familiarity with Hooke's Law and spring potential energy (U(s))
  • Knowledge of basic mechanics and energy conservation principles
  • Ability to manipulate algebraic equations for problem-solving
NEXT STEPS
  • Review the derivation of Hooke's Law and its application in spring problems
  • Study energy conservation principles in mechanical systems
  • Learn how to calculate spring constants using experimental data
  • Explore the relationship between height, compression, and velocity in spring dynamics
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding energy conservation principles in mechanical systems involving springs.

coheedcoheed
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1. Homework Statement [/b
A 2.7 kg block is dropped from rest from a height of 4.5 m above the top of the spring. When the block is momentarily at rest, the spring is compressed by 25.0 cm. What is the speed of the block when the compression of the spring is 15.0 cm?


Homework Equations


PE=mgh
KE=.5mv^2
U(s)=.5kx^2

The Attempt at a Solution



Ok so what I did at first was use PE= U(s) to find the spring constant, which i got to be 4026 N/m (I don't know if this is right, correct me if I'm wrong), after I found that, i used mgh + .5mv^2 = .5kx^2, where h was .1m, and x was .15 since it was compressed .15m. but its wrong. please help!
 
Physics news on Phys.org
1) If the v = 7.52 m/s, I would ask how you found k? I* punched few numbers and had different k (~3000)
2)Why you are taking h to be 0.1 m? in mgh + .5mv^2 = .5kx^2
 

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