Final answer:
A force F1 = (3 N)i + (4 N)j acting on a particle moving from (0, 0) to (5, 6) meters does 39 Joules of work. If the force is F1 = (2y N/m^2)i + (3x N/m)j and only moves along the x-axis, the work done is 37.5 Joules.
Step-by-step explanation:
To determine the work done by a force when an object moves from one point to another in a plane, we consider the force components and the displacement components. Work done by a force is found by calculating the dot product of the force vector and the displacement vector.
For example, consider a particle moving under the influence of a force F1 = (3 N)i + (4 N)j along a straight path from the origin (0 m, 0 m) to point (5 m, 6 m). The work done by this force can be found by calculating the dot product of the force and displacement:
W = Fx ⋅ Δx + Fy ⋅ Δy = (3 N) ⋅ (5 m) + (4 N) ⋅ (6 m) = 15 Nm + 24 Nm = 39 J.
The work done is 39 Joules. Similarly, if the force is F1 = (2y N/m2)i + (3x N/m)j, and the particle moves only in the x-direction from the origin to a point 5 meters to the right, the force in the y-direction is zero throughout the motion, and so is the work done due to that component. Therefore:
W = ∫05 Fx dx = ∫05 (3x) dx = 3/2 x2 |05 = 3/2 ⋅ 52 - 3/2 ⋅ 02 = 37.5 J.
The work done by F1 over this path is 37.5 Joules.