Conservation of energy: mass-spring system

AI Thread Summary
In a mass-spring system, a 1 kg block on a frictionless surface is connected to a spring (k=50 N/m) and a 0.45 kg dangling mass. When the mass is released, energy conservation principles apply, equating gravitational potential energy to spring potential energy. The relevant equations include U=1/2kx^2 and w=mg, which help in determining the distance the mass falls before stopping. The solution indicates that the distance fallen is approximately 0.176 m. Understanding the conservation of energy allows for solving the problem using a single variable for distance.
Avery Woodbury
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1. A 1 kg block is on a flat frictionless surface. Attached to a relaxed spring (k=50N/m). A light string is attached to the block and runs over a frictionleas pulley to a .45kg dangling mass. If the dangling mass is released from rest, how far will if fall before stopping?

Homework Equations


U= 1/2kx^2
F=-kx
W=Fd
w=mg

(I'm not exactly sure which ones I do/don't need)[/B]

The Attempt at a Solution

:

I really have no idea where to begin. The answer the book gives is .176m. [/B]
 
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Write the energy conservation law for this system, and use the only energies that are present. So for gravity you would have a gravitational potential, and due to the spring there's also a spring energy, given by the U in your relevant equations section. Due to the nature of the conservation of energy, you may equate the two energies and notice how the distances of motion that both blocks trace are the same so you can use that two have only one variable for distance - x, and ultimately solve for it.
 
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