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Math Help - A 42.0 kg block of ice

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    A 42.0 kg block of ice

    A 42.0 kg block of ice slide down a plane with an incline of 34 degree assuming friction is negligible, what is the acceleration of the block down the incline?
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    Quote Originally Posted by Refujoi
    A 42.0 kg block of ice slide down a plane with an incline of 34 degree assuming friction is negligible, what is the acceleration of the block down the incline?
    Set up your Free-Body Diagram. I am going to assume the incline is down and to the right. For a coordinate system I am going to say that +x is down the incline and +y is perpendicular to and coming out of the plane of the incline.

    There are two forces in the FBD: the weight, w, acting straight downward, and the normal force, N, acting in the +y direction.

    The only force having a component in the x direction is the weight, so:
    \sum F_x = w sin \theta = ma
    where \theta is the angle of incline. (Note that it is not cosine!)

    Since w = mg we may solve this equation for a and we get
    a = g sin( \theta ).

    (Note that the acceleration is less than g as it should be.)

    -Dan
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