Spring in a motionless elevator

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SUMMARY

The discussion focuses on calculating the spring constant for a mass m1 = 3.1 kg suspended from a spring in a motionless elevator, where the spring is extended by x = 11 cm. The relevant equation used is F = -kx, where F represents the spring force and k is the spring constant. Since the system is in equilibrium, the weight force (w = mg) equals the spring force, leading to the equation -kx = mg. This relationship allows for the determination of the spring constant k.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with Hooke's Law (F = -kx)
  • Basic knowledge of forces and equilibrium
  • Ability to perform unit conversions (e.g., cm to m)
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  • Calculate the spring constant using k = mg/x
  • Explore the implications of spring constants in different materials
  • Learn about potential energy stored in springs
  • Investigate the effects of mass and gravity on spring behavior
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Students studying physics, particularly those focusing on mechanics and spring dynamics, as well as educators looking for examples of equilibrium in systems involving springs.

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Homework Statement



A single mass m1 = 3.1 kg hangs from a spring in a motionless elevator. The spring is extended x = 11 cm from its unstretched length.
1)What is the spring constant of the spring?

Homework Equations



F=-kx


The Attempt at a Solution



Having trouble setting this one up. I know it is easy but the spring confuses me. I drew a free body diagram and I am still lost.
 
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Hi there! The only forces on the mass will be the weight force (w = mg) pulling down, and the spring force (F = -kx) pulling up. Since the mass is not accelerating, these two forces will be equal and opposite.
 
Yup, got it. -kx=mg. Durrrr. Thanks.
 

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