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Decart

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?

  • one year ago
  • one year ago

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  1. raspberryjam
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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.

    • one year ago
  2. raspberryjam
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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)

    • one year ago
  3. Decart
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    okay

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

    • one year ago
  5. raspberryjam
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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.

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

    • one year ago
  7. Decart
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    do I convert the energy into velocity

    • one year ago
  8. Decart
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    I was saying mgh as an example

    • one year ago
  9. raspberryjam
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    No just solve for v in the equation above

    • one year ago
  10. raspberryjam
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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.

    • one year ago
  11. Decart
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    that worked! I do not quite understan how though.

    • one year ago
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