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each of the four corners of a square with edge 3.00 m is occupied by a point mass 12.0 kg. there is a fifth mass, also 12.0 kg, at the center of the square. to remove the mass from the center to a point far away, how much work that must be done by an external agent?

User DZenBot
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To remove the mass from the center to a point far away, an external agent must do work equal to the change in potential energy. The work done is approximately 221.7 J.

To remove the mass from the center to a point far away, an external agent must do work equal to the change in potential energy. The potential energy of an object is given by the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object. In this case, the mass is 12.0 kg and the height is the distance from the center of the square to a point far away.

The distance from the center of the square to a point far away is the same as the length of the diagonal of the square, which can be found using the Pythagorean theorem. The diagonal of a square with side length 3.00 m is calculated as √(3.00² + 3.00²) = 3.00√2 m.

Therefore, the work that must be done by an external agent to remove the mass from the center to a point far away is equal to the change in potential energy:

W = PE = mgh = (12.0 kg)(9.8 m/s²)(3.00√2 m).

After plugging in the values and calculating, the work done by an external agent is approximately 221.7 J.

User Bwerks
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