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Spring-like trampoline

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data

    A spring-like trampoline dips down 0.08 m when a particular person stands on it. If this person jumps up to a height of 0.26 m above the top of the trampoline, how far with the trampoline compress when the person lands?

    I am very lost! Please help!
    2. Relevant equations

    F=kx

    U= .5 * k * x^2

    PE= mgh

    3. The attempt at a solution


    PE = U

    m * 9.8 * .18 = .5 * k * .18^2
     
  2. jcsd
  3. Dec 4, 2013 #2

    rock.freak667

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    Well initially when the person is standing, weight = spring force or mg = (0.08)k.

    So mg/k = 0.8

    Now in mgh = 1/2 kx ^2, you can divide by k.
     
  4. Dec 4, 2013 #3
    .08k = .5 * k * x^2

    .08= .5 x^2

    .16= x^2

    x= .40

    but it marked it wrong
     
  5. Dec 4, 2013 #4

    rock.freak667

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    Because you are equating spring force to energy

    Your equation is PE = U or mgh= ½kx2
     
  6. Dec 4, 2013 #5
    .08k *.26 = .5 * k * x^2
    .0208 = .5 * x^2
    .0146 = x^2
    x = .2039

    am I using the wrong height?

    .08k * h = .5 * k * .26^2
    .08 * h = .0338
    h = .4225

    none of these are right
     
    Last edited: Dec 4, 2013
  7. Dec 5, 2013 #6

    rock.freak667

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    Right initially the spring is compressed 0.08 m so that mg = 0.08k

    Now as the person falls, they will travel a distance 'h' to the point where they just hit the trampoline and they will continue to now compress the spring a distance 'x'. The total of these are then converted into the elastic potential energy of the spring.

    So now you have mgh + mgx = ½kx2.

    You will have a quadratic to solve in x.
     
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