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Spring Force and kinetic energy

  1. Dec 3, 2015 #1
    A novelty clock has a 0.0109 kg mass object bouncing on a spring that has a force constant of 1.34 N/m.

    How many joules of kinetic energy does the object have at its maximum velocity if the object bounces 3.49 cm above and below its equilibrium position?
     
  2. jcsd
  3. Dec 3, 2015 #2

    haruspex

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    This sounds like a homework question.
     
  4. Dec 3, 2015 #3
    It is, I think I put it in the homework forum, I didn't know that was a thing until after I posted this
     
  5. Dec 3, 2015 #4
    Ive done (.05)(force constant)(mass)^2 and I still haven't gotten the correct answer.
    Anyone know how to get it?! Please show me :(
     
  6. Dec 3, 2015 #5

    haruspex

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    Mass2? Where did that come from? This is not gravitational attraction.
     
  7. Dec 3, 2015 #6
    (.05)(k)(m)^2 isn't an equation? first I was doing (.05)(k)(m), got it wrong, and then my friend said to square the mass, however it is still wrong
     
  8. Dec 3, 2015 #7

    haruspex

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    Oh dear.
    Look up your notes, or some websites. What is the law for the force in a stretched (or compressed) spring? What is the equation for the potential energy?
     
  9. Dec 3, 2015 #8
    I know KE = (.05)(m)(v)^2
    KE = (.05)(m)(v)^2 +mgh

    Spring PE = (.05)(k)(x)^2
    x: the distance?
    k: force constant
     
  10. Dec 3, 2015 #9

    haruspex

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    Yes, much better.
    Suppose that the spring extension is A at the equilibrium position, and the amplitude of oscillation is A (=3.49cm, but keep all the working symbolic for now).
    What will be conserved as the mass rises from its lowest point to the equilibrium position? Can you write that as an equation?
     
  11. Dec 4, 2015 #10
    Those are probably just typos but you need to fix (.05) to (0.5) in both equations.
     
  12. Dec 4, 2015 #11

    haruspex

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    Well spotted.
     
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