Force acting on the particle directed towards the origin:
To show that the force acting on the particle is always directed towards the origin, we need to analyze the force vector F = ia cos(wt) + jb sin(wt).

The force vector can be separated into its x and y components:
F_x = ia cos(wt)
F_y = jb sin(wt)

The x-component of the force, F_x, is always positive and directed towards the origin. This is because the cosine function is positive in the first and fourth quadrants of the unit circle, which correspond to positive x-values. Therefore, F_x is directed towards the origin along the x-axis.

Similarly, the y-component of the force, F_y, is always negative and directed towards the origin. This is because the sine function is negative in the second and third quadrants of the unit circle, which correspond to negative y-values. Therefore, F_y is directed towards the origin along the negative y-axis.

Since both the x and y components of the force are directed towards the origin, the force vector itself is always directed towards the origin.

Force field is conservative:
To show that the force field is conservative, we need to demonstrate that the curl of the force vector is zero. The curl of a vector field is a measure of the rotation or circulation of the field.

Taking the curl of the force vector F = ia cos(wt) + jb sin(wt), we have:
∇ x F = (∂F_y/∂x - ∂F_x/∂y) k
= (0 - 0) k
= 0

Since the curl of the force vector is zero, the force field is conservative.

Work done by the force field:
To find the work done by the force field in moving the particle from point (a,0) to (0,b), we need to calculate the line integral of the force vector along the path connecting these two points.

The line integral of a vector field F along a curve C is given by:
W = ∫C F · dr

Substituting the force vector F = ia cos(wt) + jb sin(wt) and the differential displacement vector dr = dx i + dy j into the line integral formula, we have:
W = ∫C (ia cos(wt) + jb sin(wt)) · (dx i + dy j)

The path connecting the points (a,0) and (0,b) can be parameterized as:
x = at
y = bt

Taking the differentials dx = a dt and dy = b dt, the line integral becomes:
W = ∫C (ia cos(wt) + jb sin(wt)) · (a dt i + b dt j)
= ∫C (a^2 cos(wt) + b^2 sin(wt)) dt

Integrating with respect to t, the limits of integration are from 0 to 1 (as t goes from 0 to 1 when moving from (a,0) to (0,b)), we have:
W = [a^2/w sin(wt) - b^2/w cos(wt)] from 0 to 1
= (a^2/w sin(w) - b^2/w cos(w)) - (0 - b^2/w cos