A particle moves from the origin to the point x = 3.0 m, y = 15 m along the curve y =ax^2 - bx, where a = 3.0 m^{-1} and b = 4.0. It is subject to a force vector F = cxyi +dj, where c = 10 N/m^2 and d = 18 N. Calculate the work done by the force.

To calculate the work done by a force, we need to integrate the dot product of the force vector and the displacement vector along the path of the particle.

First, let's find the displacement vector. The particle moves from the origin (x = 0, y = 0) to the point (x = 3.0 m, y = 15 m). The displacement vector can be found by subtracting the initial position from the final position:

Δr = (x2 - x1) i + (y2 - y1) j
= (3.0 - 0) i + (15 - 0) j
= 3.0i + 15j

Now, let's calculate the dot product of the force vector and the displacement vector:

F · Δr = (cxyi + dj) · (3.0i + 15j)
= cxy(3.0) + d(15)
= c(3.0)(15) + d(15)

Given that c = 10 N/m² and d = 18 N, we can substitute these values into the equation:

F · Δr = (10 N/m²)(3.0)(15) + (18 N)(15)
= 450 N/m² + 270 N
= 720 N

Therefore, the work done by the force is 720 joules (J).