To compute the **work** done by the force F = (y cos z - zy sin z, ay + z^2 + z + acos a) on the object moving along the **triangle**,

we can integrate the dot product of the force and the displacement **vector** along each segment of the triangle.

The work done is given by the line integral:

Work = ∫ F · dr,

where F is the force vector and dr is the differential displacement vector.

Let's **compute** the work done along each segment of the triangle:

Segment 1: From (0,0) to (0,5)

In this segment, the **displacement** vector dr = (dx, dy) = (0, 5) and the force vector F = (y cos z - zy sin z, ay + z^2 + z + acos a).

So, the work done along this segment is:

Work1 = ∫ F · dr

= ∫ (0, 5) · (y cos z - zy sin z, ay + z^2 + z + acos a) dx

= ∫ (5y cos z - 5zy sin z, 5ay + 5z^2 + 5z + 5acos a) dx

= ∫ 0 dx + ∫ (5ay + 5z^2 + 5z + 5acos a) dx

= 0 + 5a∫ dx + 5∫ z^2 dx + 5∫ z dx + 5acos a ∫ dx

= 5a(x) + 5(xz^2) + 5(xz) + 5acos a (x) | from 0 to 0

= 5a(0) + 5(0)(z^2) + 5(0)(z) + 5acos a(0) - 5a(0) - 5(0)(0^2) - 5(0)(0) - 5acos a(0)

= 0.

So, the work done along the first segment is 0.

Segment 2: From (0,5) to (2,3)

In this segment, the displacement vector dr = (dx, dy) = (2, -2) and the force vector F = (y cos z - zy sin z, ay + z^2 + z + acos a).

So, the work done along this segment is:

Work2 = ∫ F · dr

= ∫ (2, -2) · (y cos z - zy sin z, ay + z^2 + z + acos a) dx

= ∫ (2y cos z - 2zy sin z, -2ay - 2z^2 - 2z - 2acos a) dx

= 2∫ y cos z - zy sin z dx - 2∫ ay + z^2 + z + acos a dx

= 2∫ y cos z - zy sin z dx - 2(ayx + z^2x + zx + acos ax) | from 0 to 2

= 2(2y cos z - 2zy sin z) - 2(a(2)(2) + (3)^2(2) + (2)(2) + acos a(2)) - 2(0)

= 4y cos z - 4zy sin z - 8a - 12 - 4 - 4acos a.

Segment 3: From (2,3) to (0

,0)

In this segment, the displacement vector dr = (dx, dy) = (-2, -3) and the force vector F = (y cos z - zy sin z, ay + z^2 + z + acos a).

So, the work done along this **segment** is:

Work3 = ∫ F · dr

= ∫ (-2, -3) · (y cos z - zy sin z, ay + z^2 + z + acos a) dx

= ∫ (-2y cos z + 2zy sin z, -2ay - 2z^2 - 2z - 2acos a) dx

= -2∫ y cos z - zy sin z dx - 2∫ ay + z^2 + z + acos a dx

= -2∫ y cos z - zy sin z dx - 2(ayx + z^2x + zx + acos ax) | from 2 to 0

= -2(-2y cos z + 2zy sin z) - 2(a(0)(-2) + (0)^2(-2) + (0)(-2) + acos a(0)) - 2(0)

= 4y cos z - 4zy sin z + 4acos a.

Now, we can calculate the total work done by summing the work done along each segment:

Work = Work1 + Work2 + Work3

= 0 + (4y cos z - 4zy sin z - 8a - 12 - 4 - 4acos a) + (4y cos z - 4zy sin z + 4acos a)

= 8y cos z - 8zy sin z - 8a - 20.

Therefore, the work done performed by the force F = (y cos z - zy sin z, ay + z^2 + z + acos a) on the object moving along the triangle from (0,0) to (0,5), from (0,5) to (2,3), from (2,3) to (0,0) is 8y cos z - 8zy sin z - 8a - 20.

Learn more about **vectors** here: brainly.com/question/24256726

#SPJ11