A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the bottom of the circular loop, the speed o

Question

A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the bottom of the circular loop, the speed of the bucket is 6.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the bottom of the circular loop.

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Hồng Cúc 2 weeks 2021-09-05T08:44:52+00:00 1 Answers 0 views 0

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    2021-09-05T08:46:48+00:00

    Answer: 36.0 m/s², 54.0 N, 14.7 N, 68.7 N

    Explanation:

    acceleration is centripetal

    ac = v²/r = 6.00² /1.00 = 36.0 m/s²

    The net force causes acceleration. In this case centripetal.

    Fc = Fnet = 1.50(36.0) = 54.0 N

    Fg = 1.50(9.81) = 14.7 N

    The total force supplies the necessary centripetal acceleration and supports the weight of the bucket.

    Ftot = mg + m(ac) = m(g + v²/r)

    Ftot = 1.50(9.81 + 36.0) = 68.715 ≈ 68.7 N

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