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Homework Help: Conservation of Energy with heat

  1. Nov 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A 45 kg steel ball is projected vertically with an initial speed of 280 m s . While the ball is rising, 8.5E5 J of heat energy are produced due to air friction. What is the maximum height
    reached by the ball?

    2. Relevant equations
    Ek = 1/2mv^2 Ep = mgh
    3. The attempt at a solution
    I started with Ekbefore=Epafter+Ekafter-Eheat
    so It'll be 1/2(45kg)(280m/s) = (45kg)(9.8m/s^2)h + 1/2(45kg)(0) - 8.5E5 J
    Solving for H, I got 5927.44 m, the answer is 2100m (rounded), am I suppose to add the heat energy or subtract it from the energy after?
  2. jcsd
  3. Nov 20, 2016 #2
    What do you get if you add it? This should give you your answer.
  4. Nov 20, 2016 #3
    So I'm suppose to add the heat energy to the energy after?
  5. Nov 20, 2016 #4
    Yes. Why do you think that this is so?
  6. Nov 20, 2016 #5
    It's because of conservation of energy? The energy before = the energy after
  7. Nov 20, 2016 #6
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