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An object moving in the xy-plane is subjected to the force F=(2xy +3yx )N, where x and y are in m. The particle moves from the origin to the point with coordinates (a,b) by moving first along the x-axis to (a,0), then parallel to the y-axis. How much work does the force do?

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Final Answer:

The work done by the force F as the object moves from the origin to the point (a, b) is given by the line integral of F along the path taken by the object.

Step-by-step explanation:

The work done (W) by the force F along a path can be calculated using the line integral formula:

W = ∫(2xy + 3yx) dx + ∫(2xy + 3yx) dy

where F is the force vector, dr is the differential displacement vector along the path, and the integration is performed over the path C.

In this scenario, the force is given by F = (2xy + 3yx) i + (2xy + 3yx) j, and the path is described as moving first along the x-axis to (a, 0), then parallel to the y-axis to (a, b). Let's denote the position vector along the path as r = (x, y).

The work done along the x-axis segment is calculated as:

W_x = ∫(2xy) dx from 0 to a

The work done along the y-axis segment is calculated as:

W_y = ∫(3yx) dy from 0 to b

Finally, the total work done is the sum of W_x and W_y:

W = W_x + W_y

The specific calculations involve integrating the given expressions with respect to x and y over their respective intervals.

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