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    1-calculate the work done by a long distance runner during a 10 km race. The mass of the runner is 52 kg.
    gravitational field strength is 9.8N/Kg


    2 - suggest a reason why the actual amount of work done is likely to be higher than the calculated value
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    I think you missed the force at which he went....
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    (Original post by neluxsan)
    I think you missed the force at which he went....
    I need to calculate the force first and after the work done I think is why the gave me the gravitational field strength
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    Weight acts downward with G.P.E (Gravitational Potential Energy), which is independent of the work done horizontally. This is because cos90=0.....hence when applied to W=F*D, W=0...so independent....So I think you are missing something....
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    [QUOTE=neluxsan;76478526]Weight acts downward with G.P.E (Gravitational Potential Energy), which is independent of the work done horizontally. This is because cos90=0.....hence when applied to W=F*D, W=0...so independent....So I think you are missing something....[/QUOTE/]


    that question work out of 3 marks which means that I need to convert the km into metres
    calculate the force applied and calculate the work done
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    neluxsan is correct - unless this runner is running 10km vertically upwards, then it's not the gravitational force which counts.

    Have you given us the full question?
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    (Original post by phys981)
    neluxsan is correct - unless this runner is running 10km vertically upwards, then it's not the gravitational force which counts.

    Have you given us the full question?
    yes i did
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    (Original post by chrisyabantony14)
    yes i did
    Does the question come from a book?

    In order to correctly calculate the work done, we need either the horizontal force, assuming he's running horizontally, or we have to assume he's running 10km vertically (which isn't likely as he'd run out of oxygen )

    Force and distance have to be in the same direction. if they're not, we have to take the cosine of the angle between them as described above. When force and distance moved are at right angles to each other, thereis no work done.


    If this is a question that your teacher has made up for you, be gentle when you explain this
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    Sorry, My notifications didn't pop up when you replied....but like phys981 said.....there is something wrong with your question.
 
 
 
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