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  1. #1
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    force

    A force of F = i +2j-3k is applied to a particle that moves 10 meters in the direction of i+j.

    How much is done?

    Please teach me how to solve this question. Thank you very much.
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    A force of F = i +2j-3k is applied to a particle that moves 10 meters in the direction of i+j.

    How much is done?

    Please teach me how to solve this question. Thank you very much.
    I think you mean,
    "how much work is done".

    Okay, we need to compute, the line integral,
    \int_C \bold{F}\cdot d\bold{R}
    Thus, I assume, the curve C is from,
    (0,0) to (10/\sqrt{2},10\sqrt{2},)
    (Because that is i+j, 10 units in length.)
    The parameterized curve is,
    t\bold{i}+t\bold{j}, 0\leq t\leq \sqrt{10}/2
    Thus, the line integral is,
    \int_0^{\sqrt{10}/2} 1(t)'dt+\int_0^{\sqrt{10}/2} 2(t)'dt+\int_0^{\sqrt{10}/2} 0(0)'dt
    Thus,
    \frac{3\sqrt{10}}{2}

    Physics IS NOT my thing. Do not count on the authencity of this solution.
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    Hi perfecthacker,

    Do you have a simpler way to solve this question? If yes, please teach me.
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  4. #4
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    Quote Originally Posted by Jenny20 View Post
    A force of F = i +2j-3k is applied to a particle that moves 10 meters in the direction of i+j.

    How much is done?

    Please teach me how to solve this question. Thank you very much.
    Find the component of F parallel to i+j.

    A unit vector parallel to i+j is (i+j)/sqrt(2).

    The component of F in the direction we want is then:

    F.(i+j)/sqrt(2)=(1+2)/sqrt(2)=3/sqrt(2).

    This moves 10m, so the work done is: 10 (3/sqrt(2)=30/sqrt(2).

    RonL
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jenny20 View Post
    Hi perfecthacker,

    Do you have a simpler way to solve this question? If yes, please teach me.
    Specifically, to add to CaptainBlack's (typically excellent answer) I will say that if the force is constant and the direction of motion does not vary then the work done will simply be:
    W = \vec{F} \cdot \vec{s}
    where \vec{F} is the force and \vec{s} is the displacement (the directed distance over the path taken) and \cdot is the "dot product" between the two vectors. This is the case for your problem.

    If the force is varying or the direction of motion changing then we need to use ThePerfectHacker's method.

    I would guess that you are in a Math class by the way the question was worded. If this were a Physics class I would tell you that we usually don't make a student use the integral form in an Introductory class. (There are exceptions but they are usually dealt with during class time, not homework.) What a Math class will throw at you I can't say.

    -Dan
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  6. #6
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    Quote Originally Posted by topsquark View Post
    I would guess that you are in a Math class by the way the question was worded. If this were a Physics class I would tell you that we usually don't make a student use the integral form in an Introductory class. (There are exceptions but they are usually dealt with during class time, not homework.) What a Math class will throw at you I can't say.
    One of my pet hates about Physics taught in the Maths department is -
    where did the meaning go?

    RonL

    (From one who has sat through a Maths departments idea of courses on
    QM, GR, EM&SR and would still be wondering what they were about if it
    was not for a lot a reading and a higher degree in Astrophysics)
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    One of my pet hates about Physics taught in the Maths department is -
    where did the meaning go?
    I do not understand.

    ---
    Where did I make a mistake?
    I was sure I was doing it right.
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    I do not understand.

    ---
    Where did I make a mistake?
    I was sure I was doing it right.
    Nowhere (well it must have gone wrong somewhere), but you are
    using a lot a machinery to solve a problem which can be done in a
    line or two by knowing what the concepts mean, and how they
    relate to one another.

    That is you seem more interested in the machinery than in what
    it is used for.

    RonL
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    (Referring to TPH)
    That is you seem more interested in the machinery than in what
    it is used for.
    No! Never!

    -Dan
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