Hooke's Law: Finding the Spring Constant with a 50g Mass and 7.0cm Stretch

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SUMMARY

The discussion focuses on calculating the spring constant using Hooke's Law, specifically for a spring subjected to a 50g mass that stretches an additional 7.0cm when an extra 20g is added. The correct approach involves using the formula mg = kx, where m is the total mass (70g) and x is the total stretch (7.0cm). The confusion arises from whether to consider the total mass or just the additional mass added to the spring. The accurate calculation confirms that the spring constant k can be derived from the total mass and the total elongation.

PREREQUISITES
  • Understanding of Hooke's Law and its formula (mg = kx)
  • Basic knowledge of mass and weight conversion (grams to kilograms)
  • Familiarity with unit conversions (cm to meters)
  • Ability to perform algebraic manipulations to solve for k
NEXT STEPS
  • Review the derivation of Hooke's Law and its applications in physics
  • Practice problems involving different masses and spring constants
  • Explore the relationship between mass, force, and acceleration in spring systems
  • Investigate real-world applications of Hooke's Law in engineering and design
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Students studying physics, particularly those focusing on mechanics and spring dynamics, as well as educators looking for examples of Hooke's Law applications in problem-solving.

flyingpig
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Homework Statement




A 50g mass hangs at the end of a hookean spring. When a 20g more is added to the spring, it stretches 7.0cm more. Find the spring constant.

The Attempt at a Solution



Isn't it just

mg = kx

mg/x = k

Shouldn't m = 70g instead of 20g?

My key says m = 20g?
 
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flyingpig said:
Shouldn't m = 70g instead of 20g?
That would give the total elongation from the unweighted position, not the additional elongation.
 

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