Tarzan swings on a 31.0 m long vine initially inclined at an angle of 36.0◦ with the vertical. The acceleration of gravity if 9.81 m/s 2 . W

Question

Tarzan swings on a 31.0 m long vine initially inclined at an angle of 36.0◦ with the vertical. The acceleration of gravity if 9.81 m/s 2 . What is his speed at the bottom of the swing if he a) starts from rest? Answer in units of m/s

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Bình An 5 years 2021-09-05T09:33:31+00:00 1 Answers 23 views 0

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  1. Dau
    0
    2021-09-05T09:35:11+00:00
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    Answer:

    v=10.777m/s

    Explanation:

    Tarzan swing can be thought of as change in potential energy by going from higher location We solve for height of beginning of the swing by using simple cosine equation:

    So

    31Cos36=25.08\\E_{potential}=mgh\\

    ΔE=mg(h₂-h₁)

    =m*9.81(31-25.08)\\

    The potential energy of Tarzan initial position is converted into kinetic energy of his swing.By using kinetic equation

    E_{kinectic}=P_{potential}\\1/2mv^{2}=m*9.81(31.0-25.08)\\(1/2)v^{2}=9.81(31.0-25.08)\\0.5v^{2}=58.07\\v^{2}=58.07/0.5\\v=\sqrt{58.07/0.5}\\ v=10.777m/s    

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