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Homework Help: Work/Energy question

  1. Nov 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A 0.50 kg ball starting from position A which is 7.5m above the ground, is projected down an incline as shown. Friction produces 10.7J of heat energy.

    The ball leaves the incline at position B traveling straight upward and reaches a height of 13.0m above the floor before falling back down.

    What was the initial speed, Vi, at position A (ignore air resistance)?

    This picture link that I have posted with this question is the exact same picture that I have on my question sheet except the height value at B is 13.0m not 4.00m and the height value at A is 7.5m not 3.00m.
    http://qaboard.cramster.com/answer-board/image/5b5efcf498963c65f5517ad6687243f2.jpg [Broken]

    2. Relevant equations
    ΔE = heat

    (I know that we can use the equation Wnc = ΔE, but our teacher would prefer it if we used this method to show how we got the answer)

    3. The attempt at a solution
    My question is that, I know that it says friction/heat is produced but what sign do I apply to it (positive or negative)? I put it into the equation as a positive value because it says friction is "produced", but anyway could someone tell me how you would know if it is positive or negative and if my calculations are okay?

    Since it says that friction produces 10.7J of heat energy, I know that work is not conserved in this system. Therefore I used the equation:

    ΔE = heat
    ΔEp + ΔEk = heat
    (mghf - mghi) + ((1/2)mvf(squared) - (1/2)mvi(squared)) = heat
    mghf - mghi - (1/2)mvi(squared) = heat
    (0.50)(9.80)(13.0) - (0.50(9.80)(7.5) -(1/2)(0.50)vi(squared) = 10.7
    63.7 - 36.75 -(1/2)(0.50)vi(squared) = 10.7
    26.95 -10.7 = (1/2)(0.50)vi(squared)
    (2)(16.25) = (0.50)vi(squared)
    32.5/0.50 = vi(squared)
    √65 = √(vi(squared)
    8.062257748 = vi
    8.06 m/s = vi
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 3, 2008 #2


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    Homework Helper

    Hi Inertialforce! :smile:

    Write the equation in the form:

    energy before = energy after ± …

    you know it's lost energy, so which side must be larger? :wink:

    (sorry, i didn't chcek your equations, they're too difficult to read :redface:)
  4. Nov 3, 2008 #3
    so, energy is lost therefore the right side has to equal a negative number? Which makes the left side larger meaning that either the change in potential energy is larger than the change in kinetic energy or vice versa. Would this be correct?
  5. Nov 3, 2008 #4


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    I think your confusing yourself by looking at it that way …

    you're certainly confusing me :redface:

    Just say KE + PE before = KE + PE after + energy lost. :smile:
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