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Thread: Gravity-Work Question

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    Gravity-Work Question

    How much work is done by gravity in causing a 24 kg rock to tumble 50m down a slope at an angle 48 degrees to the verticle, to one decimal place?
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by peon123 View Post
    How much work is done by gravity in causing a 24 kg rock to tumble 50m down a slope at an angle 48 degrees to the verticle, to one decimal place?
    I believe it is $\displaystyle W = Fscos\theta$

    $\displaystyle F = mg$
    $\displaystyle s = 50$
    $\displaystyle cos\theta = cos(42)$ (this works for the horizontal angle)

    $\displaystyle W = 24 \times 9.81 \times cos(42) \times 50 = 8748,3J = 8.7kJ$

    I think that's right, if you have an answer please post it because there could be another way to do it
    Last edited by e^(i*pi); May 29th 2009 at 02:22 PM. Reason: see post 5
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    Super Member craig's Avatar
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    Quote Originally Posted by peon123 View Post
    How much work is done by gravity in causing a 24 kg rock to tumble 50m down a slope at an angle 48 degrees to the verticle, to one decimal place?
    Work done is equal to $\displaystyle Fs$, where $\displaystyle F$ is the force acting on the particle and $\displaystyle s$ is the displacement.

    If I remember my mechanics correctly, then as the question is asking you the work done by gravity, it is the horizontal displacement that you are after, which is equal to $\displaystyle 50 \sin{48}$.

    Then as we are interested in the direction of gravity, would it be $\displaystyle 24g.50 \sin{48}$?
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    Super Member craig's Avatar
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    This works out to be $\displaystyle 8739.4J$ or $\displaystyle 8.7kJ$.

    Just to note, I have taken $\displaystyle g$ as 9.8, as opposed to 9.81.
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    Super Member craig's Avatar
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    Just realised that we used the same equations

    Quote Originally Posted by e^(i*pi) View Post
    $\displaystyle W = 24 \times 9.81 \times cos(42) \times 50 = 87483J ~ 8.7kJ$
    There's a typo in your answer, it should be $\displaystyle 8748.3J$, not $\displaystyle 87483J$
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by craig View Post
    Just realised that we used the same equations



    There's a typo in your answer, it should be $\displaystyle 8748.3J$, not $\displaystyle 87483J$
    Yeah, I checked that on my calculator, makes sense since $\displaystyle sin(90-x) = cos(x)$

    Thanks for that, I worked it out in kJ first but had to expand for the 1dp
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