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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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The  centripetal acceleration of the bucket at the bottom of the circular loop is 36 m/s².

The net force at the bottom of the circular loop is 68.7 N.

The given parameters:

  • Mass of the bucket, m = 1.5 kg
  • Radius of the circle, r = 1 m
  • Speed of the bucket at bottom, v = 6 m/s

What is centripetal acceleration?

  • Centripetal acceleration is the radial acceleration of an object in a circular path.

The  centripetal acceleration of the bucket at the bottom of the circular loop is calculated as follows;

[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{6^2}{1} \\\\a_c = 36 \ m/s^2[/tex]

The net force at the bottom of the circular loop is calculated as follows;

[tex]F_{net} = m(g + a_c)\\\\F_{net} = 1.5(9.8 + 36)\\\\F_{net} = 68.7 \ N[/tex]

Learn more about centripetal acceleration here: https://brainly.com/question/79801

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