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anonymous

  • 5 years ago

When a particle is located a distance x meters from the origin, a force of 2cos(pix/6) newtons acts on it. How much work W is done in moving the particle from x = 1 to x = 5?

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  1. anonymous
    • 5 years ago
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    Is this calculus based physics?

  2. anonymous
    • 5 years ago
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    I will assume that this is a calculus based physics question. W = F * D Here the force is variable, so: integral 2cos(pix/6) dx from lower limit 1 to upper limit 5.

  3. anonymous
    • 5 years ago
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    right, i got stumped because of the negative force. what does that mean? doesnt force have to be positive

  4. anonymous
    • 5 years ago
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    whats the positive and negative direction here?

  5. anonymous
    • 5 years ago
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    A force can always work against a system in which you ascribe a vector to be positive.

  6. anonymous
    • 5 years ago
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    Does that make sense?

  7. anonymous
    • 5 years ago
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    so what is positive and what is negative

  8. anonymous
    • 5 years ago
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    can you make an example

  9. anonymous
    • 5 years ago
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    The most simple example would probably be gravity.

  10. anonymous
    • 5 years ago
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    Don't let semantics play with your mind, you can designate whichever direction you choose as postive or negative.

  11. anonymous
    • 5 years ago
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    so a particle along the x axis , or x meters from the origin experiences a force 2cos (pix/6). what direction is positive and what is negative. it can be up or down?

  12. anonymous
    • 5 years ago
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    *cos

  13. anonymous
    • 5 years ago
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    but positive and negative are like east /west or north south, right?

  14. anonymous
    • 5 years ago
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    yes

  15. anonymous
    • 5 years ago
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    graph the function that you have, look at interval x=1 to x=5 and you'll understand how you can have a negative force here.

  16. anonymous
    • 5 years ago
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    yes it oscillates

  17. anonymous
    • 5 years ago
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    you are finding the area under that curve, and there is more area under the x-axis than above.

  18. anonymous
    • 5 years ago
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    right its symmetric and adds up to zero

  19. anonymous
    • 5 years ago
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    yes, sorry, you're right it's zero, I was just trying to explain how it could be negative. does it make sense now? sometimes, it's really helpful to look at a graph.

  20. anonymous
    • 5 years ago
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    a force is just a vector, it can be negative, because it has magnitude & direction*

  21. anonymous
    • 5 years ago
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    ok thanks, can i ask a quick question

  22. anonymous
    • 5 years ago
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    i just want to reiterate whats going on

  23. anonymous
    • 5 years ago
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    crap my browser froze

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