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    sunshineangel's Avatar
    sunshineangel Posts: 43, Reputation: 7
    Junior Member
     
    #1

    Nov 15, 2008, 10:53 AM
    Finding Work in Physics
    If I'm trying to figure out work and I am given efficiency how does that factor into the equation when I already know the weight, gravity, and change in height.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #2

    Nov 15, 2008, 06:27 PM

    The efficiency simply multiplies the total energy. A process that releases 24Joules, but extracting that energy is only 50% efficient, then the actual energy useful for work is just 24*0.5 = 12J.

    Ask if you have further questions! :D
    sunshineangel's Avatar
    sunshineangel Posts: 43, Reputation: 7
    Junior Member
     
    #3

    Nov 15, 2008, 07:10 PM

    Okay then if you have the actual work as being 12J than is ideal work 24J? And is that the same thing as actual and ideal mechanical advantage?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #4

    Nov 16, 2008, 04:31 AM

    Sort of, yes, Actual mechanical advantage takes things into account such as bearing friction and air resistance etc. Ideal mechanical advantage ignores these (it is "ideal")
    sunshineangel's Avatar
    sunshineangel Posts: 43, Reputation: 7
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    #5

    Nov 16, 2008, 11:46 AM

    That makes total sense, thanks you have helped on what started as a very frustrating project.

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