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Kinetic and Potential Energy

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data

    A 1.50 m high trampoline stores 50.0J or elastic potential energy as a 39.2Kg child lands on it. What is the maximum speed the child will reach as he is propelled vertically into the air and how high above the ground will he reach before he lands again?

    2. Relevant equations

    Not sure what it means :/

    Ek = Ep
    Ek=1/2*m*v^2
    Ep=m*g*h

    Ek= kinetic energy
    Ep= potential energy
    m=mass
    v=speed
    h=height
    g=gravitational force (9.81m/s^2)


    3. The attempt at a solution

    v= [Square Root]((Ek)/(1/2)(m))
    v= [Square Root]((50.0J)/(0.5)(39.2Kg))
    v= 1.5971 m/s
    v= 1.60 m/s <---- Maximum speed i think


    h= (Ep)/(m*g)
    h= (50.0J)/(39.2Kg)(9.81m/s^2)
    h= 0.13002 m
    h= 0.130 m <---- Maximum height i think
     
  2. jcsd
  3. Sep 30, 2011 #2
    That's almost correct. Be careful on your height. Is 0.130m the maximum height reached by the child?

    Another thing: the solution is numerically correct, but I think that you should state clearly which are the types of energy (kinetic, potential gravitational, potential elastic...) in this problem, and how they are linked, e.g. why the kinetic energy Ek is 50.0J?
    I know this may seem quite boring because it is obvious, but this way of thinking will help you in more difficult exercises.
     
  4. Oct 1, 2011 #3
    I found something wrong but I'm still not sure if the answer is correct.

    Edit:
    h= (Ep)/(m*g)
    h= (50.0J)/(39.2Kg)(9.81m/s^2)
    h= 0.13002 m
    h= 0.130 m + 1.50m (because above ground)
    h= 1.63m <----Possible answer
    ----------------------------------------------
    I tried stating why Kinetic=Potential but I still couldn't figure out the answer. :(

    Mechanical Energy= Kinetic Energy+Potential Gravitational Energy
    so
    [Change in]Kinetic Energy = [Change in]Potential Gravitational Energy

    Please help, I just learned about energy so I'm not really good at it
     
  5. Oct 2, 2011 #4
    You got it.
    The best way is:
    Choose the reference to measure the potential gravitational energy; I will choose the level of the trampoline (with this convention, h=0 means on the trampoline) , but you can choose ground level as well.

    At the beginning: Only Potential elastic energy [itex]E_{elastic}=50.0J[/itex] (because I choose to put Egravitational=0 on the trampoline
    While flying: Potential gravitational energy +Kinetic energy: [itex]E_{gravit}+K[/itex]

    Since you must have
    [itex]E_{elastic}=E_{gravit}+K=mgh+\frac{1}{2}mv^2[/itex]
    to get the highest speed you have to solve in v, to get the highest height you solve on h.
    [itex]v_{max}=\sqrt{\frac{2[E_{elastic}-mgh]}{m}}[/itex] the maximum speed is when h=0, and you get [itex]v_{max}=\sqrt{\frac{2E_{elastic}}{m}}[/itex].
    [itex]h_{max}=\frac{E_{elastic}-\frac{1}{2}mv^2}{mg}[/itex] which has its maximum when v=0, so that
    [itex]h_{max}=\frac{E_{elastic}}{mg}[/itex]

    Got it?
     
  6. Oct 2, 2011 #5
    thanks i get it now :)
     
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