Mechanics - Hooke's law and energy conservation

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SUMMARY

The discussion focuses on applying Hooke's law and energy conservation principles to determine the maximum depth a bead will fall when attached to an elastic string. The participant initially used incorrect equations related to modulus of elasticity instead of the correct stiffness constant, leading to confusion in deriving the final equation. The correct total energy equation is established as Et = (mgx^2)/(2l), which incorporates gravitational potential energy and elastic potential energy. The participant is encouraged to streamline their four-step solution into a more cohesive approach for clarity.

PREREQUISITES
  • Understanding of Hooke's Law and elastic potential energy
  • Familiarity with energy conservation principles in physics
  • Knowledge of kinematic equations (suvat) for motion analysis
  • Ability to manipulate algebraic equations for problem-solving
NEXT STEPS
  • Study the derivation of Hooke's Law and its applications in mechanics
  • Learn about energy conservation in elastic systems
  • Practice solving problems involving kinematic equations and energy transformations
  • Explore advanced topics in dynamics, such as oscillations and wave motion
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of energy conservation and elastic properties in real-world applications.

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Homework Statement


One end of a light elastic string of stiffness mg/l and natural length l is attached to a point O. A small bead of mass m is fixed to the free end of the string. The bead is held at O and then released so that it will fall vertically. In terms of find the greatest depth to which it will fall below O.

Homework Equations


Ek=(mv^2)/2 Es=(ex^2)/(2l) Eg=mgh suvat

The Attempt at a Solution


Now i started off by splitting the motion up into 4 parts. Part 1 before it dropped Et = Eg so total energy is mgh which is mg(l+x). Part 2 is as its fell a distance l Et = Eg + Ek. Using suvat i got the speed so i got the equation Et = xmg + mlg. Part 3 will be taken at any time while the mass is moving and extending the string. Part 4 is at the maximum extension and not moving so i got Et=Es which is Et = (mgx^2)/2.

To me that all seems correct but when i try combining the equations to get x i can't seem to get anything that works. Can someone show me where I've gone wrong please? Thanks

I've just noticed I've been using the wrong equations and using modulus of elasticity not k so the final equation should be Et= (mgx^2)2l
 
Last edited:
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You could try to patch up your four-step solution or think about how to reduce it to one step.
 

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