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Work done by varying force

  1. Nov 30, 2012 #1
    1. The problem statement, all variables and given/known data
    calculate the work done by a force F=(4.00yj)N(a varying force) that acts on an object that moves along the Y-axis, from Y=-2.0m to Y=3.0m


    2. Relevant equations
    non given but i assume it is W=fscos(θ)


    3. The attempt at a solution
    attempted to use W=fscos(θ) but only made me more lost. i have no idea where to even begin
     
  2. jcsd
  3. Nov 30, 2012 #2

    SteamKing

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    Re: the mindblower

    What's the direction of F as it moves from y = -2 to y = 3?

    Is the direction of F the same as the path from y = -2 to y = 3?
     
  4. Nov 30, 2012 #3
    Re: the mindblower

    I would assume so, because it is acting on the object that is moving, so technically it is the force causing the movement is moving in the same direction. But no other info was given.
     
  5. Nov 30, 2012 #4
    Re: the mindblower

    Start from the actual definition of work, [tex] W = \int_\gamma d\vec{s} \cdot \vec{F} [/tex] where you integrate over the path of the object.
     
  6. Nov 30, 2012 #5
    Re: the mindblower

    no clue on how to do it this way.
     
  7. Nov 30, 2012 #6
    Re: the mindblower

    You know how to integrate and do dot products, right?

    You have [itex] d\vec{s} = dy \hat{j} [/itex] (as the object moves on y-axis) and [itex]\vec{F} = 4 y N \hat{j} [/itex].
     
  8. Nov 30, 2012 #7

    HallsofIvy

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    Re: the mindblower

    You don't have to "assume" so that information is expressly given:
    "calculate the work done by a force F=(4.00yj)N(a varying force) that acts on an object that moves along the Y-axis, from Y=-2.0m to Y=3.0m"

    That "j" in the force is the unit vector in the x direction. That means that the [itex]cos(\theta)[/itex] in your "W=fscos(θ)" is irrelevant. [itex]\theta= 0[/itex] so [itex]cos(\theta)= 1[/itex]. But just "W= fs" is wrong because that is for constant force.

    Now, have you taken or are you taking a Calculus class? Clearly whoever gave you this problem expects you to know that with a variable force, that "product", fs, becomes an integral: [itex]\int f(x)dx[/itex].
     
  9. Nov 30, 2012 #8
    Re: the mindblower

    i did not take calculus, this is a "stepup problem" according to our professor to get us ready for calc based physics.
     
  10. Nov 30, 2012 #9

    vela

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    Re: the mindblower

    Try using the fact that the work is equal to the area under the curve when you plot the force vs. y.
     
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