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Homework Help: Conservation of energy of a block of ice

  1. Feb 10, 2008 #1
    Well... It is first first time on the Physics Forums, and I'm stuck on this problem:

    1. A 2.5 kg block of ice at a temperature of 0.0 °C and an initial speed of 5.7 m/s slides across a level floor. if 3.3 × 10^5 J are required to melt 1.0 kg of ice, how much ice melts, assuming that the initial kinetic energy of the ice block is entirely converted to the ice's internal energy?

    2. ∆PE + ∆KE +∆U = 0
    (the change in potential energy + the change in kinetic energy + the change in internal energy = 0)
    PE = mgh
    KE = ½mv²

    3. Well... I really have no idea how to solve this one...

    But it sounds to me that KE = ½(2.5 kg)(5.7 m/s)² somewhere in there... But I really don't know what I am supposed to solve for (I think it is the mass of how much ice melts). I don't know how to integrate 3.3 × 10^5 J into an equation. If I could get a equation to start off from, I could do the rest. Thanks.
    Last edited: Feb 10, 2008
  2. jcsd
  3. Feb 10, 2008 #2


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    Are you sure the problem doesn't give you a *final* velocity? The basic concept is that while the block is sliding across the floor, friction does negative work on it, reducing its kinetic energy. The kinetic energy lost heats up the block (i.e. it is converted into internal energy). If you knew the final speed, you'd know the change in kinetic energy. From that, you could calculate how much ice melts.
  4. Feb 10, 2008 #3
    No final velocity is given... Is there a way I can find it from that information?

    EDIT: 3.3 × 10^5 J is circled in my textbook for some reason...
    Last edited: Feb 10, 2008
  5. Feb 10, 2008 #4


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    Yeah, maybe we are meant to interpret "the initial kinetic energy is ENTIRELY converted..." as meaning all of it. In that case, the final kinetic energy would be zero.
  6. Feb 10, 2008 #5
    All right, I'll attempt to work off of that. :)

    Here we go:

    PEi + KEi +Ui = PEf + KEf + Uf

    PEi & PEf = 0

    KEf = 0

    KEi +Ui = Uf

    KEi = Uf - Ui <--- (∆U)

    ½mv² = ∆U

    ½(2.5 kg)(5.7 m/s)² = ∆U

    42 J = ∆U

    3.3 × 10^5 J/1.0 kg = 42 J/m

    m(3.3 × 10^5 J) = (42 J)(1.0 kg)

    m = 42 J(kg)/3.3 × 10^5 J

    m = 1.3 × 10^-4 kg

    1.3 × 10^-4 kg of ice melts...

    That's not much, isn't it? I hope this is correct...
    Last edited: Feb 10, 2008
  7. Feb 10, 2008 #6
    Does it seem that I got the right answer?
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