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Thread: [SOLVED] Work

  1. #1
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    [SOLVED] Work

    I was able to get this question correct but now I can't.

    Let F be a constant unit force that is parallel to the vector <-1,0,1> in xyz-space. What is the work done by F on a particle that moves along the path given by (t,t^2,t^3) between time t=0 and time t=1?

    W=cos(\theta)\parallel F\parallel \parallel PQ\parallel

    cos(\theta)=\frac{<,u,v>}{\parallel u\parallel \parallel v\parallel}

    simplifies down to:
    W=<u,v>=<F,PQ>

    For PQ, I have (but something is wrong since the answer is 0) 0\leq x\leq 1, t\mathbf{i}+t^2\mathbf{j}+xt^3\mathbf{k}

    W=-t+t^3\neq 0
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  2. #2
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    You're using the wrong expression for Work. It should be the line integral \int_C {\bf F}\cdot d {\bf r}.
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  3. #3
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    Quote Originally Posted by ojones View Post
    You're using the wrong expression for Work. It should be the line integral \int_C {\bf F}\cdot d {\bf r}.
    I don't remember how to setup the integral with what is giving.

    F is the force, C is the path, and ds is ??

    Since this path is in 3-space, I thinking this will be an iterated integral?
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  4. #4
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    Buy a calculus book and read it.
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  5. #5
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    Quote Originally Posted by ojones View Post
    Buy a calculus book and read it.
    Really? Is that how it is done? I thought information was disseminate through osmosis.
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  6. #6
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    F=(-1,0,1)
    r(t)=(t,t^2,t^3)
    F(r(t))=(-1,0,1)
    r'(t)=(1,2t,3t^2)
    F(r(t))r'(t)=-1+3t^2
    \int F(r(t))r'(t) \: dt=\int (-1+3t^2) \: dt \: = \: -t+t^3 \: |_0^1 \: = \: 0
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  7. #7
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    Correct.
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