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Normal force work homework

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Starting from rest, a 5.0kg block slides 2.5m down a rough 30 degree incline in 2.0s. Determine the following:
    A) the work done by the forces of gravity ( my answer 61.31J)
    B) the mechanical energy lost due to friction ( my answer 45.69 J)
    A) the work done by the normal force between the block and the incline ( i need help )


    2. Relevant equations
    W=fdcos(0)
    Fn=mg


    3. The attempt at a solution
    I tried F(2.5m)cos30)
    15.63J=2.165F
    F=7.219N
     
  2. jcsd
  3. Jan 10, 2010 #2
    Re: Work

    i found out a and b . i dont know c like you
     
  4. Jan 10, 2010 #3
    Re: Work

    The normal force is not simply mg. Remember, your block sits on an incline, and the normal force is always perpendicular to the plane!
     
  5. Jan 10, 2010 #4

    Matterwave

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    Re: Work

    If jeagues' hint wasn't clear enough, what's the work done on an object by a force that is perpendicular to the direction of motion of that object?
     
  6. Jan 10, 2010 #5
    Re: Work

    F= Fgp-Ff
    F=Fsin30-uFn
     
  7. Jan 10, 2010 #6

    Matterwave

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    Re: Work

    Let's forget about the incline. Say I just a block that is moving horizontally on the floor, what's the work the normal force is doing?
     
  8. Jan 10, 2010 #7
    Re: Work

    What is [tex] \vec{N} \cdot d\vec{r} [/tex]?
    [tex] \vec{N} [/tex] being the normal force and [tex] d \vec{r}[/tex] being the direction of the blocks motion.
     
  9. Jan 10, 2010 #8
    Re: Work

    Apply this suggestion to a simple context if you're still having trouble understanding.

    What if we had a ball rolling along a level floor, again, what's the work done by a force that is perpendicular to the direction of motion?

    HINT: W = fdcos(theta)
     
  10. Jan 10, 2010 #9
    Re: Work

    W=fdcos(0) so

    W=f x 2.5m cos 180
     
  11. Jan 10, 2010 #10

    Matterwave

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    Re: Work

    no, theta in this case is 90 not 0
     
  12. Jan 10, 2010 #11
    Re: Work

    You need to review your understanding of the word perpendicular.
     
  13. Jan 10, 2010 #12
    Re: Work

    i know what perpendicular is, i dont get what you're asking

    wouldn't it be Fgp-Ff
     
  14. Jan 10, 2010 #13

    Matterwave

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    Re: Work

    What's W=f*d*cos(90)?
     
  15. Jan 10, 2010 #14
    Re: Work

    If you think about the case I described with the ball rolling along a horizontal floor the direction of motion is simply horizontal. So perpendicular to that would give theta an angle of 90.

    Now in the case of your inclined plane, the way you've done your work so far you've defined the x axis parallel to the plane and the y axis perpendicular to the plane. So if you again look at the angle theta down your plane (with respect you the axis's you've been using) you will see that the angle will again be 90.
     
  16. Jan 10, 2010 #15
    Re: Work

    it will always be 0 because cos 90 is always 0
     
  17. Jan 10, 2010 #16
    Re: Work

    Do you know why physically the work is 0?
     
  18. Jan 10, 2010 #17
    Re: Work

    no clue
     
  19. Jan 10, 2010 #18
    Re: Work

    im guessing it's because the way the box only moves up and down the ramp
     
  20. Jan 10, 2010 #19
    Re: Work

    The box is moving down the ramp.
    The normal force is perpendicular to the direction of motion.
    The normal force plays no role in assisting or hindering the MOTION of the box.
    Thus the force can't do any work. The Normal Force is not pushing the box over any distance.

    I hope that makes sense.
     
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