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Changing temperature of falling bodies

  1. Jul 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A block of material of mass m and specific heat c falls from height h and reaches speed v just before striking the ground. Its temperature is measured immediately after it strikes the ground. If we ignore any change in temperature owing to interaction with the air, the change in temperature of the block of material is

    a. v2/2c
    b. gh/c
    c. vgh/c
    d. All of the answers above are correct.
    e. Only (a) and (b) above are correct

    2. Relevant equations

    mgh = deltaQ + .5mv2

    Q = mcT

    3. The attempt at a solution

    I'm confused because by my understanding, the answer should be (gh-(v2/2))/c which is a combination of both A and B. However, this does not mean that both A and B are correct as answer choice E states. I obtained this answer by simply isolating T in the aforementioned relative equation. Am I missing something?
  2. jcsd
  3. Jul 18, 2009 #2


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    When the block is at height h it has potential energy mgh and kinetic energy 0. Just as it hits the ground it has kinetic energy (1/2)*mv^2 and potential energy 0. Conservation of energy says KE+PE is constant. That means (1/2)*mv^2=mgh. Your answer would give 0. Do you see why both a) and b) are correct now?
  4. Jul 18, 2009 #3
    Ah, i must have read the question too quickly. Earlier in the homework assignment there was a question that involved an object reaching terminal velocity and then continued falling, allowing for further temperature change. I assumed it had the same conditions as the earlier problem. Thanks!
  5. Jul 18, 2009 #4
    I am slightly confused on one thing though... Since KE = PE, then wouldn't deltaQ be zero, so there wouldn't be any temperature change at all?
  6. Jul 18, 2009 #5


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    What actually happens here is PE (mgh) at the top of the fall get changed into an equal amount of KE (mv^2/2) at the impact which then gets changed into an equal amount of heat energy (Q) conserving energy all the way. They are all EQUAL. How would that translate into 'no temperature change'??
  7. Jul 18, 2009 #6
    Alright I understand it now. Thank you.
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