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Math Help - work

  1. #1
    Junior Member winterwyrm's Avatar
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    work

    The only force acting on a 3.8 kg canister that is moving in an xy plane has a magnitude of 4.7 N. The canister initially has a velocity of 4.2 m/s in the positive x direction, and some time later has a velocity of 4.0 m/s in the positive y direction. How much work is done on the canister by the 4.7 N force during this time?

    Note: I'm sure it has something to do with the work-kinetic energy theorum. thx
    Last edited by winterwyrm; October 4th 2008 at 12:29 AM.
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  2. #2
    Member Greengoblin's Avatar
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    Work is given by W=Fs where F = force and s = the distance over which the force is applied (for a changing force it's an integral).

    So you need to find the distance. First find the acceleration using F=ma.
    <br />
a=\frac{F}{m} = \frac{4.7}{3.8} =

    Then using equations of motion, v^2=u^2+2as you can get the distance and then the work. (v is final velocity, u is initialvelocity, a is acceleratio, and s is the distance)

    You should be able to do the calculations.
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  3. #3
    MHF Contributor
    skeeter's Avatar
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    since F = 4.7 N is the only force acting on the object, it does all all the work which happens to also be the net work on the object.

    so, yes, it has everything to do with the work-KE theorem.

    W_{net} = \Delta KE = \frac{1}{2}m(v_f^2 - v_0^2)
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  4. #4
    Junior Member winterwyrm's Avatar
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    ... Oh no, these two methods give two different results... Or maybe I'm just messing something up...
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