91Ó°ÊÓ

In physics, a force acting in two dimensions, or a 2D force, means that the force has components in both the x and y directions. For instance, consider the force vector in the given problem: \( \vec{{F}} = ((2 \text{{N/m}}) x) \hat{{i}} + (3 \text{{N}}) \hat{{j}} \). This vector tells us that:

• The force in the x-direction, denoted by \hat{{i}}, changes with the position \((x)\). It is \((2 \text{{N/m}}) x \), which means the force grows stronger as x increases.
• In contrast, the force in the y-direction, denoted by \hat{{j}}, is a constant \(3 \text{{N}}\). It doesn't change with the position.

In this manner, we individually account for how forces act in each dimension. Understanding this separation is crucial for solving problems involving multi-dimensional forces.

The displacement vector represents the change in position of a particle from one point to another. It is determined by subtracting the initial position vector from the final position vector. Given \( \vec{r}_1 = (2 \text{{ m}}) \hat{i} + (3 \text{{ m}}) \hat{j} \) and \( \vec{r}_2 = -(4 \text{{ m}}) \hat{i} - (3 \text{{ m}}) \hat{j} \), the displacement vector \( \Delta \vec{r} \) can be found as: \( \Delta \vec{r} = \vec{r}_2 - \vec{r}_1 \) Substituting the given values, we get: \( \Delta \vec{r} = \left(-(4 \text{{ m}}) \hat{i} - (3 \text{{ m}}) \hat{j} \right) - \left((2 \text{{ m}}) \hat{i} + (3 \text{{ m}}) \hat{j} \right) = (-6 \text{{ m}}) \hat{i} - (6 \text{{ m}}) \hat{j} \). This results in a displacement vector of \( \Delta \vec{r} = (-6 \text{{ m}}) \hat{i} - (6 \text{{ m}}) \hat{j} \), effectively capturing the distance and direction moved in each dimension.

To determine the work done by a force, particularly when it varies with position, we use the integral of force. Work is a measure of energy transfer and, in this context, is computed by integrating the force over the displacement. For the x-direction, the work done \( W_x \) is calculated by: \( W_x = \int_{x_1}^{x_2} F_x \cdot dx \) Given that \( F_x = (2 \text{{ N/m}}) x \), the integral can be expressed as: \( W_x = \int_{2}^{-4} (2 \text{{ N/m}}) x \; dx \). Solving this integral, we get: \( W_x = 2 \left[ \frac{x^2}{2} \right]_{2}^{-4} = \left[ x^2 \right]_{2}^{-4} = (-4)^2 - (2)^2 = 16 - 4 = 12 \text{ J} \). For the y-direction, where the force is constant: \( W_y = F_y \cdot \Delta y \) with \( F_y = 3 \text{{ N}} \) and the displacement is \( \Delta y = y_2 - y_1 = -3 \text{{ m}} - 3 \text{{ m}} = -6 \text{{ m}} \), So, \( W_y = 3 \text{{ N}} \cdot (-6 \text{{ m}}) = -18 \text{{ J}} \). Summing up the work done in both directions, the total work done \( W \) is: \( W = W_x + W_y = 12 \text{{ J}} + (-18 \text{{ J}}) = -6 \text{{ J}} \). Integrals help account for variable forces, enabling precise calculations of work done over a displacement.