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Thermal power problem

  1. Oct 13, 2006 #1
    Hello, here is the problem Ive been stuck on. It seems simple but I am getting the wrong answer. Any help would be appreciated.


    A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200.

    What average thermal power is produced as the rock stops?


    My answers thus far are
    784
    1040
    427

    They were all wrong.
     
  2. jcsd
  3. Oct 13, 2006 #2

    Andrew Mason

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    You are right. They are all wrong. Besides, you don't have any units. What are the units of power here?

    Perhaps you could explain your thinking first. How did you analyse the problem?

    AM
     
  4. Oct 13, 2006 #3

    OlderDan

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    How do you think this problem should be done? Your wrong answers do not tell us what you are thinking.
     
  5. Oct 13, 2006 #4
    Thanks Andrew for responding.

    It should be in W.

    for DeltaX I did ((1/2)(.2)(9.8*20kg))/8ms that got me the distance 2.45m

    Then I solved for acceleration and got 13.0612

    Then I solved for force which was 13.0612*20kg = 261.224

    I also solved for time which was .6125s

    so I did V average of 2.45m/.6125s and got 4ms

    So then I had everything I needed. to solve for Power average

    P_av= F*V_av

    So it was 261.224*4ms = 1044.9

    But it said that was wrong??? I dont know why.


    Thanks, again for responding
     
  6. Oct 13, 2006 #5
    Thanks you guys for responding to my question. I hope we can figure it out.
     
  7. Oct 13, 2006 #6
    Okay, I solved it.

    it was : ((mg)(friction)(x))/(time)

    ((9.80)(20)(.2)(2.45)/(.6125s) = 156.8 W
     
  8. Oct 13, 2006 #7

    Andrew Mason

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    I am not sure how you determined x and time. One way to look at it is:

    [tex]P_{avg} = Fv_{avg} = \mu_k mg\Delta v/2 = .2*20*9.8*4 = 156.8 W[/tex]

    AM
     
  9. Oct 16, 2007 #8
    i dont get it...
    F=mg
    V[tex]_{}ave[/tex]= v/2

    but where did the [tex]\mu[/tex] came from?
     
  10. Oct 16, 2007 #9
    what equations did manipulate to arrive at this one [tex]P_{avg} = Fv_{avg} = \mu_k mg\Delta v/2 = .2*20*9.8*4 = 156.8 W[/tex]
     
  11. Mar 5, 2010 #10
    F is not mg. F = umg or in other words the force of friction.
     
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