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Kinetic Energy - Force done by friction

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Kelli weighs 435 N, and she is sitting on a playground swing seat that hangs 0.44 m above the ground. Tom pulls the swing back and releases it when the seat is 1.00 m above the ground. Assume that air resistance is negligible.

    (a) How fast is Kelli moving when the swing passes through its lowest position?
    SOLVED: 3.31 m/s

    (b) If Kelli moves through the lowest point at 2.0 m/s, how much work was done on the swing by friction?
    SOLVED: 154.284 Joules

    V at bottom in closed system = 3.31 m/s
    V at bottom in open system = 2.0 m/s
    Kelli's mass = 44.388 kg
    K in closed system = 243.158 Joules

    2. Relevant equations

    K = .5mv2 = J(oules)

    W = Fd = J(oules)

    3. The attempt at a solution

    I attempted to use the difference of the two velocities to find the kinetic energy in Joules, and it did NOT work.

    Any suggestions?
    Last edited: Dec 11, 2008
  2. jcsd
  3. Dec 11, 2008 #2
    Not 100% sure but did this class today
    For b) you have to do conservation of energy .5mv^2+mgh+(work done by friction)= 0. From there you should have enough informationto solve for (work done by friction).
  4. Dec 11, 2008 #3
    Which height should I use for the "change in potential energy part" or the formula? 1.00 m? .44 m? or .56 m?

    I tried all three, and none gave me the answer I need.
    Last edited: Dec 11, 2008
  5. Dec 11, 2008 #4
    You should use .56 since the bottom of the swing is your refrence point
  6. Dec 11, 2008 #5
    I now know the answer to part (b) is 154. 824 J, obtained by using the equation :


    Can anyone explain how this give me the answer? I'm aware that [tex]\left(F\times\left(1-d\right)\right)[/tex] is the formula for Work. How does [tex]-\left(\frac{2F}{g}\right)[/tex] apply?
  7. Dec 11, 2008 #6


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    Homework Helper

    No need to do all that.

    You know m*g*h = 435*(.56) = 243.6 J
    In part a) that yields your v that I presume you calculated correctly.

    In part b) they give you what the actual KE was so the Frictional losses must be the difference between the 2.

    Your mass is 435/9.8 = 44.38 kg which yields 88.776 J of KE.

    PE - KE = Work by friction.
  8. Dec 11, 2008 #7
    "PE - KE = work by friction" makes so much sense now that you said it.

    Thanks a bunch. :D
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