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How much energy is dissipated in the fourth bounce?

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  1. Oct 26, 2016 #1
    1. The problem statement, all variables and given/known data
    A 0.70-kg basketball dropped on a hardwood floor rises back up to 66 % of its original height.If the basketball is dropped from a height of 1.4 m , how much energy is dissipated in the first bounce?( answer =3.3J).
    How much energy is dissipated in the fourth bounce?

    2. Relevant equations
    U=mgy

    3. The attempt at a solution
    i know that at the top of the first bounce the kinetic energy is zero and the potential energy is 9.604J. After the first bounce theres 6.304 of the potential energy left, meaning 3.3 J was dissipated. I tried using the same approach as in part one, assuming its losing 66% of the height each bounce since i am not told otherwise.
    i got, a height of 0.2656m after the fourth bounce. plugging that into the formula, U=mgy i got
    U=(0.7kg)(9.8m/s2)(0.2656m)
    =1.8J
    then i subtracted that from the original potential energy
    9.604-1.8= 7.8J
    i got the answer of 7.8J of energy was dissipated. this is incorrect and i am unsure as to why
     
  2. jcsd
  3. Oct 26, 2016 #2
    i caught my mistake it was looking just from the third bounce potential energy to the fourth bounce, not the system as a whole
     
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