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Decart

  • 2 years ago

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?

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  1. raspberryjam
    • 2 years ago
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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.

  2. raspberryjam
    • 2 years ago
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    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)

  3. Decart
    • 2 years ago
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    okay

  4. Decart
    • 2 years ago
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    i thought the potential would be at total displacement like mgh

  5. raspberryjam
    • 2 years ago
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    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.

  6. raspberryjam
    • 2 years ago
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    Also you can't use mgh in this case because it's a spring-based problem...Hooke's law

  7. Decart
    • 2 years ago
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    do I convert the energy into velocity

  8. Decart
    • 2 years ago
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    I was saying mgh as an example

  9. raspberryjam
    • 2 years ago
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    No just solve for v in the equation above

  10. raspberryjam
    • 2 years ago
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    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.

  11. Decart
    • 2 years ago
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    that worked! I do not quite understan how though.

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