91Ó°ÊÓ

In this exercise, we need to determine the displacement vector, which describes the change in position of a particle. The displacement vector \(\textbf{d}\) is calculated by subtracting the initial position from the final position. Here, the initial point is \(2, 5\) and the final point is \(7, 3\). Subtract the coordinates of the initial point from the final point: \( \textbf{d} = (7 - 2) \textbf{i} + (3 - 5) \textbf{j}\). After performing the subtraction, we get \( \textbf{d} = 5 \textbf{i} - 2 \textbf{j}\). This vector tells us that the particle has moved 5 units to the right and 2 units down.

The next step in the exercise is to find the resultant force acting on the particle. The resultant force is found by adding the individual forces together. Here, we have two forces: \( F_{1} = 3 \textbf{i} - \textbf{j} \) and \( F_{2} = -4 \textbf{i} + 5 \textbf{j} \). Adding these vectors, we get: \( \textbf{F} = (3 \textbf{i} - \textbf{j}) + (-4 \textbf{i} + 5 \textbf{j})\). To simplify, combine the \( \textbf{i} \) components and the \(\textbf{j}\) components separately. This results in \( \textbf{F} = - \textbf{i} + 4 \textbf{j} \). The resultant force vector \( \textbf{F} \) indicates that there is a net force of 1 unit to the left and 4 units upwards acting on the particle.

Finally, to find the work done by the forces, we need to compute the dot product of the resultant force vector and the displacement vector. The work done \( W \) is given by: \( W = \textbf{F} \bullet \textbf{d} \). Here, \( \textbf{F} = - \textbf{i} + 4 \textbf{j} \) and \( \textbf{d} = 5 \textbf{i} - 2 \textbf{j} \). The dot product of two vectors \( \textbf{A} \) and \( \textbf{B} \) is calculated as: \( \textbf{A} \bullet \textbf{B} = A_{i}B_{i} + A_{j}B_{j} \). Apply this to the given vectors: \( W = (-1)(5) + (4)(-2) = -5 - 8 = -13 \). Hence, the work done by the two forces is \(-13\) foot-pounds. The negative sign indicates that the forces are doing work against the direction of the displacement.