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Hooke's Law and Work Energy THeorem

  1. Nov 21, 2012 #1
    1. The problem statement, all variables and given/known data

    At a waterpark, sleds with riders are sent along a slippery,
    horizontal surface by the release of a large compressed spring. The
    spring with force constant and negligible mass
    rests on the frictionless horizontal surface. One end is in contact
    with a stationary wall. A sled and rider with total mass 70.0 kg are
    pushed against the other end, compressing the spring 0.375 m. The
    sled is then released with zero initial velocity. What is the sled’s
    speed when the spring (a) returns to its uncompressed length and
    (b) is still compressed 0.200 m?

    2. Relevant equations

    Using Work Energy THeorem


    3. The attempt at a solution

    For part b) I attemted the equation ... .5 K (x)^2 = 0.5 m (v^2) and i got an answer of 1.5 but the book attempt was different... they used it with the x of A and got the answer of 2.4 m/s ... What my problem?? Kindly Clarify... THANKS A LOT!
     
  2. jcsd
  3. Nov 21, 2012 #2

    Doc Al

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    Staff: Mentor

    Use conservation of energy. What's the total energy at any point?

    (Setting PE = KE only works in certain cases--such as when all the PE is converted to KE.)
     
  4. Nov 21, 2012 #3
    But i applied it in the first case, and i got the correct answer???.. I mean what's wrong with the approach as long as there isn't any friction or resistance??
     
  5. Nov 21, 2012 #4

    Doc Al

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    Staff: Mentor

    It applied in the first case because all the PE was transformed to KE.

    Conservation of energy will apply in all cases.

    Starting point: Spring fully compressed. What's the spring PE? What's the KE? What's the total energy?

    Ending point: Spring compressed by some amount. What's the spring PE? What's the KE? What's the total energy?
     
  6. Nov 21, 2012 #5
    Allright, can u clarify it using work energy thorem without reffering to potential energy...
     
  7. Nov 21, 2012 #6

    Doc Al

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    Staff: Mentor

    Work done by the spring = ΔKE of the mass.
     
  8. Nov 21, 2012 #7
    Great... So work done by force spring to move to the 0.2 is integral of fx dx from 0 to 0.2 which must equalise the difference in kinetic energy where k initial is zero???
     
  9. Nov 21, 2012 #8

    Doc Al

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    Staff: Mentor

    Almost. The spring moves from x = 0.375 to 0.2 .
     
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