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Thread: Change in Gravitational Potential Energy (Theta Involved)

  1. #1
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    Change in Gravitational Potential Energy (Theta Involved)

    A 100kg man walks up a steep hill that is inclined at $\displaystyle 30\theta$ to the horizontal. After walking $\displaystyle 1.0km$ along this incline, his change in gravitational potential energy will be:

    a) $\displaystyle 9.8e5 J$
    b) $\displaystyle 8.5e5 J$
    c) $\displaystyle 4.9e5 J$ Correct Answer
    d) $\displaystyle 6.0e5 J$

    Given
    $\displaystyle m = 100kg$
    $\displaystyle L_{I} = 1km = 1000m$
    $\displaystyle 30\theta$ to horizontal

    Required
    $\displaystyle E_{G}$

    Analysis

    I'm not exactly sure how to solve this with the given.
    Do I have to use the equation $\displaystyle \Delta{E_{G}} = \frac{1}{2}mv^2$?
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    As I can "guess" from your question theta here stands for degree itself

    Reason and Explanation to your answer:

    Change in gravitational potential energy is = m*g*h

    Where h is the distance above the initial line ,in your case its horizontal -
    ---------------------------------------
    Distance of man above horizontal --after he has climbed 1000m along the slope of 30degree inclination = 1000 sin(30)

    Put the
    value of g and m in the equation

    your answer

    = 100 * 1000 sin(30) * 9.8 Joule

    = 10^5 * (1/2) * 9.8 J

    =4.9 * 10^5 J

    = 4.9e5 J

    Tell me if you still need help
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