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Work and kentic energy

  1. Jul 17, 2006 #1
    this proplem is very tricky and i was wondering if anyone could help me.

    A worker pushed a 27 kg block 10.4 m along a level floor at constant speed with a force directed 25° below the horizontal.
    (a) If the coefficient of kinetic friction is 0.20, how much work was done by the worker's force?
    ?J
    (b) What was the increase in thermal energy of the block-floor system? ?J

    I decided to try using two equations. f=mewN/cos25 AND F=MG-n/SIN 25 . I equated them to fine part a but that answer did not work. any help
     
  2. jcsd
  3. Jul 17, 2006 #2


    What is the energy balance in this problem ?
     
    Last edited: Jul 17, 2006
  4. Jul 17, 2006 #3

    Doc Al

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    Is "f" meant to be the friction? If so, that first equation is incorrect.
    Is "F" meant to be the worker's force? If so, that second equation has an incorrect sign.
     
  5. Jul 17, 2006 #4
    ok i got two equations sumation of F_x=mgcos30-Force of kentic friction and f=m*mew*n cos 30.
     
  6. Jul 17, 2006 #5

    Doc Al

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    Kinetic friction is simply:
    [tex]f = \mu N[/tex]
     
  7. Jul 17, 2006 #6
    w=fdcos(theata) right
     
  8. Jul 17, 2006 #7
    anyone else know something that can be done with this problem
     
  9. Jul 17, 2006 #8

    Doc Al

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    It's certainly true that the work done by the force F is:
    [tex]W = Fd \cos\theta[/tex]
     
  10. Jul 18, 2006 #9
    anyone know something that I should do.
     
  11. Jul 18, 2006 #10

    Doc Al

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    Why don't you solve for [itex]F[/itex] or [itex]F\cos\theta[/itex] so you can calculate the work done? Hint: The block is moving at a constant speed.
     
  12. Jul 18, 2006 #11
    i have already found the normal force
     
  13. Jul 18, 2006 #12
    i used two equations F=u(k)N/cos(25) and F=(mg-N)/sin25

    I solved them to get a normal force of 242.028.

    I then subsitied the first equation F=u(k)N/cos(25) in for the F in
    w=fdcos25. I found 503 and that was wrong. I have no idea what i did wrong.
     
  14. Jul 18, 2006 #13

    Doc Al

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    This one looks OK.
    This one does not.
     
  15. Jul 18, 2006 #14
    what should i change. should the N be +
     
  16. Jul 18, 2006 #15

    Doc Al

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    Yes. But you should know why it's wrong and needs to be changed.
     
  17. Jul 18, 2006 #16
    so it should me (mg+n)/sin25 because the normal force pushes off the ground
     
  18. Jul 18, 2006 #17
    ok i got part a correct but i do not kow how to do part b. for a i got 603.8
     
  19. Jul 18, 2006 #18

    Doc Al

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    What happens to the work done by the worker? Where does the energy go?
     
  20. Jul 18, 2006 #19
    it turns into heat
     
  21. Jul 18, 2006 #20

    Doc Al

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    So what's the answer to part b?
     
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