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abannavong Group Title

Physics help with friction! the problem is "You push a 25kg wooden box across a wooden floor at a constant velocity of 1.0 m/s. If the coefficient of friction is .20, how much force do you exert on the box?"

  • one year ago
  • one year ago

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  1. abannavong Group Title
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    how do you set up the question?

    • one year ago
  2. abannavong Group Title
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    I know that m= 25kg

    • one year ago
  3. Sea15era Group Title
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    F=\[nuN\]

    • one year ago
  4. Shadowys Group Title
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    First, you draw the free body diagram. And note that the question tells you: there is no acceleration(constant velocity) Then you apply Newton's second law to it.

    • one year ago
  5. abannavong Group Title
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    ok so newtons second law you would use f=ma?

    • one year ago
  6. Shadowys Group Title
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    yes. \(\Sigma F = m\Sigma a\)

    • one year ago
  7. abannavong Group Title
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    oh ok a

    • one year ago
  8. abannavong Group Title
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    so we know that m=25 kg and what is the a?

    • one year ago
  9. Sea15era Group Title
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    |dw:1352779234119:dw|

    • one year ago
  10. Sea15era Group Title
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    first your normal force equals your mass times gravity (245N) and then plug them into your equation to find a. a=.20x245/25

    • one year ago
  11. Shadowys Group Title
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    well, since it is constant velocity, a=0.

    • one year ago
  12. Sea15era Group Title
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    for your a you should get 1.96 m/s^2

    • one year ago
  13. abannavong Group Title
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    oh ok

    • one year ago
  14. abannavong Group Title
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    so how do you get 1.96 m/s^2

    • one year ago
  15. Shadowys Group Title
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    Hmm... @Sea15era why is there acceleration from the total force?

    • one year ago
  16. Sea15era Group Title
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    |dw:1352779486046:dw|

    • one year ago
  17. Shadowys Group Title
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    but the box is moving with constant velocity, i.e. no acceleration, right?

    • one year ago
  18. Sea15era Group Title
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    thats just a velocity acceleration is in m/s^2 not m/s i believe.

    • one year ago
  19. Sea15era Group Title
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    F=25x1.96 F=49N should be the answer if i did my equations and sum of forces statement wrong.

    • one year ago
  20. abannavong Group Title
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    oh ok

    • one year ago
  21. Sea15era Group Title
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    I hope this helps you , good luck and if by chance it's wrong I'm sorry!

    • one year ago
  22. Shadowys Group Title
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    Yes. The acceleration is not given, but since you're pushing it in constant velocity, there should not be any acceleration.

    • one year ago
  23. abannavong Group Title
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    so how do you find the force that is being exerted?

    • one year ago
  24. Shadowys Group Title
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    \(\Sigma F=ma\) Start with this picture. |dw:1352781084312:dw| As shown, the forces are F_x and F_f, of different directions. Applying the second law, \(F_x - F_f =ma=0\) as it is constant velocity.

    • one year ago
  25. Shadowys Group Title
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    Now, sea told you \(F_f=\mu F_N\), where \(F_N=F_g\) in this case.|dw:1352781249983:dw| Do you follow?

    • one year ago
  26. abannavong Group Title
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    kinda

    • one year ago
  27. abannavong Group Title
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    im still a bit confused about the Fn= Fg

    • one year ago
  28. Shadowys Group Title
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    Now, as \(F_g = m g\), where m=25kg and g=9.8, so you get \(F_g\) Now after you get \(F_g\) you can get \(F_N\) After you get \(F_N\), You get \(F_f\) Then since \( F_x-F_f=0\), \(F_x=F_f\).

    • one year ago
  29. Shadowys Group Title
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    On the y-axis, v=0,(it doesn't move vertically), so a=0 too. Since \(\Sigma F_y=F_N-F_g= 0\),it follows that \(F_N=F_g\)

    • one year ago
  30. abannavong Group Title
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    oh ok i got you so far

    • one year ago
  31. Shadowys Group Title
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    Step by step, have you got the value of \(F_N\)?

    • one year ago
  32. abannavong Group Title
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    im still confused on how to get Fn

    • one year ago
  33. abannavong Group Title
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    i got Fg= 245

    • one year ago
  34. Shadowys Group Title
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    yup, and that's the value of Fn. Fn is the normal force of the ground acting on the box.

    • one year ago
  35. abannavong Group Title
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    oh ok

    • one year ago
  36. Shadowys Group Title
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    yeah, so next you will get the value of friction force...

    • one year ago
  37. abannavong Group Title
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    oh ok so how do you find the friction force

    • one year ago
  38. Shadowys Group Title
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    \(F_f = \mu F_N\) where \(\mu\) is the coefficient of kinetic friction.

    • one year ago
  39. abannavong Group Title
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    oh ok

    • one year ago
  40. abannavong Group Title
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    (.20)(25) ?

    • one year ago
  41. abannavong Group Title
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    nvm i mean (245)(.2)

    • one year ago
  42. Shadowys Group Title
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    yup. Then after you get friction force, as you see from posts above, the value of friction force is the force exerted.

    • one year ago
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