Determine Min Work & Final Velocity of Masses in Massless Spring Exp

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SUMMARY

The discussion focuses on determining the minimum work required to compress a massless spring between a 1 kg mass and a 3 kg mass, with the 3 kg mass moving at 10 m/s after release. The minimum work needed to compress the spring is established as 150 Joules, which corresponds to the energy stored in the spring. Additionally, when both masses are released simultaneously, the conservation of momentum principle is applied to find the final velocities of each mass, leading to neat integer results.

PREREQUISITES
  • Understanding of kinetic energy (KE) and its formula: KE = 1/2 mv²
  • Knowledge of spring potential energy and Hooke's Law: PE = 1/2 kx²
  • Familiarity with the conservation of momentum principle
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study the principles of conservation of energy in mechanical systems
  • Learn about Hooke's Law and its applications in spring mechanics
  • Explore momentum conservation in elastic and inelastic collisions
  • Investigate the relationship between work done on a system and energy transfer
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Physics students, educators, and anyone interested in mechanics, particularly those studying spring dynamics and energy conservation principles.

edud8
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1981B2. A massless spring is between a 1 kilogram mass and a 3 kilogram mass as shown above, but is not attached to either mass. Both masses are on a horizontal frictionless table. In an experiment, the 1 kilogram mass is held in place and the spring is compressed by pushing on the 3 kilogram mass. The 3 kilogram mass is then released and moves off with a speed of 10 meters per second.
a. Determine the minimum work needed to compress the spring in this experiment.





The spring is compressed again exactly as above, but this time both masses are released simultaneously.
b. Determine the final velocity of each mass relative to the table after the masses are released.



attempt:
1/2mv^2=1/2kx^2
1/2mv^2=150, but idk how to solve for k or x
 
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edud8 said:
1/2mv^2=1/2kx^2
1/2mv^2=150, but idk how to solve for k or x
You don't need to. You only need to know the amount of energy stored in the spring, which is the 150 J you found.

This is the total KE of both masses in part 2, and you also need to remember something else that is conserved.

Your answer will be a neat integer.
 

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