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Finding Work Done By Friction

  1. Jan 6, 2008 #1
    [SOLVED] Finding Work Done By Friction

    A 1.2g feather is dropped from a height of 1.5m and immediately reaches terminal velocity. If the time to fall is 0.95 s and force of the air resistance on the feather is constant, how much work was done on the feather by friction? Answer: W = -0.016 J

    I use the equation W = U + K which turns into W = Uf - Ui + Kf - Ki.
    Then Uf + Kf = Ui + Ki + W. I know I have to bring the W to the other side but I don't know why. Does anyone have an explanation?

    Then the equation becomes Ui = Kf + W because there's no Ki and Uf.

    Then substitute mgh for Ui and 1/2mv^2 for Kf which makes the equation:
    W = mgh - 1/2mv^2 but the answer turns out to be positive rather negative. I switch the mgh and 1/2 mv^2 and get a slightly different number it it's right. What am I doing wrong?
     
    Last edited: Jan 6, 2008
  2. jcsd
  3. Jan 6, 2008 #2

    olgranpappy

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    how sure are you that the above answer is correct?
     
  4. Jan 6, 2008 #3
    I'm very sure that it's correct since we've been doing problems since the beginning of the school year and the answer given has always been correct except once. Is it not possible for work to be a negative number?
     
  5. Jan 6, 2008 #4

    olgranpappy

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    it is possible for work to be negative... and you should be able to tell right away whether your answer for the work should be positive of negative--are the force doing the work and the displacement in the same direction or in opposite directions?

    also, what I meant in my previous post is: are you sure that you wrote the correct number of decimal places. Are you sure you didn't mean: "-0.016" instead of "-0.0016".
     
  6. Jan 6, 2008 #5
    Oh, it's suppose to be -0.016. Just fixed it in the original post. I don't understand why I had to switch the mgh and the 1/2mv^2 in order for it to be negative. Also, why did I have to move the W to the other side?
     
  7. Jan 6, 2008 #6
    You use this equation right away. No need for further manipulation. That will only make you swap minus signs.
     
  8. Jan 6, 2008 #7

    olgranpappy

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    You didn't "have to", nor should you have, but you did anyways... and got the wrong answer... and then you undid what you did to get the right answer.
     
  9. Jan 6, 2008 #8
    Oh that works but can you explain why I had to go through that entire process for this problem:

    A 35 kg school child climbs a 4.5 meter high spiral slide. If the child reaches a speed of 3.3 m/s at the bottom of the slide, how much work was done by friction?

    I tried your method on this problem and it didn't work because the equation is suppose to be W = mgh -1/2mv^2.
     
  10. Jan 6, 2008 #9

    olgranpappy

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    wrong.
     
  11. Jan 6, 2008 #10
    Wow! I can't believe I teacher did the example problem incorrectly.
     
  12. Jan 6, 2008 #11

    olgranpappy

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    Those who can't do, teach. And those who can't teach, teach gym. :wink:
     
  13. Dec 7, 2010 #12
    I'm working on the same problem but I'm having trouble, I can't quite get the right equation...I've been using the work equation and KE but I'm not sure that is right. And I don't get the equation above...
     
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