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Work friction problem

  1. Nov 12, 2004 #1

    tony873004

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    Figure 8-16: http://www.webassign.net/walker/08-16.gif

    12. [Walker2 8.P.002.] Calculate the work done by friction as a 2.1 kg box is slid along a floor from point A to point B in Figure 8-16 along paths 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.21.

    Since work can have a negative number, I'm assuming that all 3 answers will be the same, since the displacement is 3 meters. Path 1 travels 4 - 1 = 3 meters, Path 1 travels 2 + 1 = 3 meters, and Path 3 travels 3 meters. The vertical travels will not be affected by friction.

    W = f * d
    d = 3
    f = friction
    friction = mu * N (force Normal)
    N = ma
    N = 2.1 * 9.81
    N = 20.601
    friction = 0.21 * 20.601 = 4.32621
    W = 4.32621 * 3
    W = 12.98 j

    but this is wrong, and so is -12.98.

    Thanks in advance for your input :biggrin:
     
  2. jcsd
  3. Nov 12, 2004 #2
    Work is done in the direction of an applied force. In this case you can't neglect the vertical lines. Work is done on them too. You should add them up.
     
  4. Nov 12, 2004 #3

    tony873004

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    But is work done on them by friction? All it asks for is the work done by friction. And friction is mu * N, but in the vertical direction doesn't N = 0? meaning that work done by friction would also be 0? I can see why there would be work done by gravity, but we have to ignore that. I think :confused:

    In the previous problem, we had to figure out work done by gravity, but not friction and I got it right:
    figure: http://www.webassign.net/walker/08-15.gif

    [Walker2 8.P.001.] Calculate the work done by gravity as a 1.7 kg object is moved from point A to point B in Figure 8-15 along paths 1, 2, and 3.
    The answers are -33j for all 3 parts.
     
    Last edited: Nov 12, 2004
  5. Nov 12, 2004 #4
    This question is different, it is actually a plan. You look the movement of the object from above. The object is always on the floor.
     
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