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abannavong

  • 2 years ago

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?"

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  1. abannavong
    • 2 years ago
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    how do you set up the question?

  2. abannavong
    • 2 years ago
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    I know that m= 25kg

  3. Sea15era
    • 2 years ago
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    F=\[nuN\]

  4. Shadowys
    • 2 years ago
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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.

  5. abannavong
    • 2 years ago
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    ok so newtons second law you would use f=ma?

  6. Shadowys
    • 2 years ago
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    yes. \(\Sigma F = m\Sigma a\)

  7. abannavong
    • 2 years ago
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    oh ok a

  8. abannavong
    • 2 years ago
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    so we know that m=25 kg and what is the a?

  9. Sea15era
    • 2 years ago
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    |dw:1352779234119:dw|

  10. Sea15era
    • 2 years ago
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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

  11. Shadowys
    • 2 years ago
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    well, since it is constant velocity, a=0.

  12. Sea15era
    • 2 years ago
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    for your a you should get 1.96 m/s^2

  13. abannavong
    • 2 years ago
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    oh ok

  14. abannavong
    • 2 years ago
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    so how do you get 1.96 m/s^2

  15. Shadowys
    • 2 years ago
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    Hmm... @Sea15era why is there acceleration from the total force?

  16. Sea15era
    • 2 years ago
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    |dw:1352779486046:dw|

  17. Shadowys
    • 2 years ago
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    but the box is moving with constant velocity, i.e. no acceleration, right?

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

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

  20. abannavong
    • 2 years ago
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    oh ok

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

  22. Shadowys
    • 2 years ago
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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.

  23. abannavong
    • 2 years ago
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    so how do you find the force that is being exerted?

  24. Shadowys
    • 2 years ago
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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.

  25. Shadowys
    • 2 years ago
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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?

  26. abannavong
    • 2 years ago
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    kinda

  27. abannavong
    • 2 years ago
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    im still a bit confused about the Fn= Fg

  28. Shadowys
    • 2 years ago
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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 \( F_x-F_f=0\), \(F_x=F_f\).

  29. Shadowys
    • 2 years ago
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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 \(F_N=F_g\)

  30. abannavong
    • 2 years ago
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    oh ok i got you so far

  31. Shadowys
    • 2 years ago
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    Step by step, have you got the value of \(F_N\)?

  32. abannavong
    • 2 years ago
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    im still confused on how to get Fn

  33. abannavong
    • 2 years ago
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    i got Fg= 245

  34. Shadowys
    • 2 years ago
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    yup, and that's the value of Fn. Fn is the normal force of the ground acting on the box.

  35. abannavong
    • 2 years ago
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    oh ok

  36. Shadowys
    • 2 years ago
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    yeah, so next you will get the value of friction force...

  37. abannavong
    • 2 years ago
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    oh ok so how do you find the friction force

  38. Shadowys
    • 2 years ago
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    \(F_f = \mu F_N\) where \(\mu\) is the coefficient of kinetic friction.

  39. abannavong
    • 2 years ago
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    oh ok

  40. abannavong
    • 2 years ago
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    (.20)(25) ?

  41. abannavong
    • 2 years ago
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    nvm i mean (245)(.2)

  42. Shadowys
    • 2 years ago
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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.

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