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Finding work given mass and mew?

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  1. Dec 17, 2014 #1
    The problem reads:

    "someone is pulling a 200kg sleigh across a field at a constant velocity. If the coefficient of kinetic friction is .13, how much work do they do on the sleigh if they pull it 94 meters?"

    I know I have to find force net and then use the formula for work (force x distance)

    how would I go about finding force net? Force friction+Force normalizing (which would be mass x gravity?)

    any guidance would be helpful asap. thanks!
     
  2. jcsd
  3. Dec 17, 2014 #2

    Bystander

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    The way the problem is written you can assume it's a level (horizontal) field.
     
  4. Dec 17, 2014 #3

    Dick

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    You don't do any work on the normal force. It's perpendicular to the direction of displacement. Think vectors.
     
  5. Dec 17, 2014 #4
    Sorry I forgot to mention the problem also states the surface is level, and the sleigh is moving with a constant acceleration. Would I begin by finding force friction and force normalizing?
     
  6. Dec 17, 2014 #5

    Bystander

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    You'll want the normal force first.
     
  7. Dec 17, 2014 #6

    Dick

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    Yes, do that. But then tell me why only the frictional force is important for the work done.
     
  8. Dec 17, 2014 #7
    So I multiplied

    200kg x 9.8 =1960 (normal force)

    1960 x .13 = 254.8 (Force friction)

    Would these two be added together to equal force net?

    or do I just use the force friction to find the work because the surface is level?
     
  9. Dec 17, 2014 #8

    Bystander

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    Are you lifting the sled, or dragging it?
     
  10. Dec 17, 2014 #9
    pulling, there is no angle. It is a level surface
     
  11. Dec 17, 2014 #10

    Dick

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    The forces would be added together as vectors. And the equation for the work is the dot product of the force with the displacement. You do know about vectors, right?
     
  12. Dec 17, 2014 #11
    so work =force x distance

    so 2214 N x 94 m =208,116...

    this number seems to big. what am I doing wrong?

    isn't it something along the lines of setting the two equal to 0?
     
  13. Dec 17, 2014 #12

    Dick

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    You don't add forces pointing in different directions together to get a single number. You have to think about vectors. Each of those forces contributes to the work differently because the work depends on the direction of the force and the direction of the displacement. Not just the magnitude of each.
     
    Last edited: Dec 18, 2014
  14. Dec 18, 2014 #13

    Dick

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    Yes! Why not include the normal force? Tell me why.
     
  15. Dec 18, 2014 #14
    is it because the normal force is acting on the object both up and down? the surface is opposing 1960 N and the object is exerting 1960 N of force on it? so one is positive, one is negative, and they cancel? The i'd get the answer by multiplying the force friction by the distance?
     
  16. Dec 18, 2014 #15

    Dick

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    That's actually a pretty good guess. But it isn't quite complete. If a force is acting on an object perpendicular to the direction of motion then no work is done. Work isn't as simple as magnitude of force*magnitude of distance. There are directions to think about. Do you about vectors and dot products?
     
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