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A 0.321-kg mass is attatched to a spring with force constant of 13.3 N/m. If the mass is displaced 0.256 m from the Equalibrium and released, what is its speed when it is 0.128 m from equalibrium?

Physics
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\[ \frac{1}{2}mv^2=\frac{1}{2}kx_{total}^2-\frac{1}{2}kx^2\] where k is the spring constant and \[ x_{total}\] is the total displacement from equilibrium. \[x \] is the distance 0.128 m.
Law of conservation of energy yo. Kinetic energy = (potential energy of spring at total displacement) - (potential energy of spring at whatever point you're trying to find the velocity at)
okay

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Other answers:

i thought the potential would be at total displacement like mgh
Well you're trying to find the velocity at a specific point so you have to subtract it from the potential energy of when the spring is fully stretched which gives you the amt of potential energy transformed into kinetic energy.
Also you can't use mgh in this case because it's a spring-based problem...Hooke's law
do I convert the energy into velocity
I was saying mgh as an example
No just solve for v in the equation above
You can't just do total displacement because you're trying to find the velocity at a specific point which is when x=0.128 m. You aren't finding the displacement when x=0.256 m.
that worked! I do not quite understan how though.

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