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Mercker

In a tractor-pull competition, a tractor applies a force of 1.3kN to the sled, which has mass 1.1x 10^4 kg. At that point, the coefficient of kinetic friction between the sled and the ground has increased to 0.80. What is the acceleration of the sled? What is the significance of the sign of the acceleration?

  • 2 years ago
  • 2 years ago

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  1. hosein
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    |dw:1328974050914:dw| i get right along + so \[F-f=ma \] \[f=\mu _{k}mg\] \[a=(F-\mu _{k}mg)/m\]

    • 2 years ago
  2. Mercker
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    Im sorry, I dont understand what you wrote down for the formulas Is that f= Mu - overall mass? and what does the other one say?

    • 2 years ago
  3. hosein
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    you can't understand formula ok?

    • 2 years ago
  4. hosein
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    when F exerted to body, friction so exerted force on it (here not equal)

    • 2 years ago
  5. Mercker
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    Are they not equal by units? :S So far what I've done is calculated F(g), which was 1.1x10^4 X 9.81m/s^2, which got me 107910 for Fg Then because FN is assumed to be equal to Fg, I plugged in FN to Ff, which got me 107910 x .80 which got me 86328 I then did FNet= Fapplied - Ffriction, which got me -85028 Plugging that into F=ma, whiuch got me -85028= 1.1x 10^4 (x) Which yielded -7.73 m/s THe answer in the textbook is 0.61m/s Could you tell me what Im doing wrong in my steps? Im sorry for not understanding what you wrote because I still only know the very basics :(

    • 2 years ago
  6. hosein
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    M is overal minus a means that slid going to stop but i distracted ! my solve equal yours !

    • 2 years ago
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