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ksaimouli Group Title

A block is pulled by a string that makes an angle of 25º to the horizontal. If the mass of the block is 12.0 kg and the coefficient of friction is 0.25, what force would keep the block moving at a constant velocity?

  • one year ago
  • one year ago

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  1. henpen Group Title
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    |dw:1349633059599:dw|

    • one year ago
  2. ksaimouli Group Title
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    |dw:1349633142335:dw|

    • one year ago
  3. ksaimouli Group Title
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    what to do after this i am totally confused

    • one year ago
  4. ksaimouli Group Title
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    i know that ff=u*fn

    • one year ago
  5. ksaimouli Group Title
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    we need to find fn

    • one year ago
  6. henpen Group Title
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    \[F_n=mg-F_y=12*9.8-Fsin(25)\]

    • one year ago
  7. ksaimouli Group Title
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    i got this but how do u find the F

    • one year ago
  8. henpen Group Title
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    \[F_f= \mu F_n= 0.25(117.6-Fsin(25))=F_x=Fcos(25)\]

    • one year ago
  9. ksaimouli Group Title
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    how did u get 117.6

    • one year ago
  10. henpen Group Title
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    mg

    • one year ago
  11. henpen Group Title
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    12*9.8

    • one year ago
  12. henpen Group Title
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    \[29.4-0.25\sin(25)F=Fcos(25)\] \[\frac{29.4}{\cos(25)+0.25\sin(25)}=F\]

    • one year ago
  13. ksaimouli Group Title
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    thx

    • one year ago
  14. ksaimouli Group Title
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    why =fcos(25)

    • one year ago
  15. henpen Group Title
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    |dw:1349634909926:dw|Basic trig

    • one year ago
  16. henpen Group Title
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    \[F_f=F_x=Fcos(25)\]

    • one year ago
  17. henpen Group Title
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    No acceleration => no net force in x-direction

    • one year ago
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