Solving for Spring Compression: 2.3kg Dropped 4.82m

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SUMMARY

The discussion focuses on calculating the compression of a vertical spring when a mass of 2.3 kg is dropped from a height of 4.82 meters. Using the spring constant of 20 N/cm, the correct compression of the spring is determined to be 34.1 cm. The key equations utilized include potential energy (PE = mgh), kinetic energy (KE = (1/2)m(v^2)), and Hooke's Law (Hooke's Law = (1/2)K(x^2)). A common mistake highlighted is the importance of maintaining consistent units throughout the calculations.

PREREQUISITES
  • Understanding of gravitational potential energy (PE = mgh)
  • Familiarity with kinetic energy equations (KE = (1/2)m(v^2))
  • Knowledge of Hooke's Law and spring constants
  • Ability to convert units (N/cm to N/m)
NEXT STEPS
  • Review the principles of energy conservation in mechanical systems
  • Learn about unit conversion techniques in physics problems
  • Explore advanced applications of Hooke's Law in different contexts
  • Investigate the effects of varying spring constants on compression outcomes
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to energy and spring dynamics.

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Homework Statement


A mass of 2.3 kg is dropped from a height of 4.82 meters above a vertical spring anchored at its lower end to the floor. If the spring constant is 20 N/cm, how far, to the nearest tenth of a cm, is the spring compressed?
answer is 34.1

Homework Equations


PE = mgh
KE = (1/2)m(v^2)
Hookes Law= (1/2)K(x^2)

The Attempt at a Solution


mgh = (1/2)K(x^2)
 
Last edited:
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There you go, that is how you solve the equation. The actual solution, however, is wrong. Did you keep your units?
 

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