91Ó°ÊÓ

A force vector is a representation of a force that includes both magnitude and direction. In physics, forces are often represented as vectors because they can operate in multiple dimensions and directions.
Vectors are denoted by an arrow over the variable, such as \(\vec{F}\). A vector has components along different axes, and these components are labeled with unit vectors, usually \(\hat{i}\) for the x-axis and \(\hat{j}\) for the y-axis.
In this exercise, the force vector depends on position: \(\vec{F}=3x\hat{i}+4y\hat{j}\). This means that the force's magnitude and direction vary with the object's position.

• The 3x term indicates a component of the force in the direction of the x-axis that increases with x.
• The 4y term indicates a component of the force in the direction of the y-axis that grows with y.

Understanding force vectors is crucial when determining how an object will move or change due to applied forces.

Differential work refers to the small amount of work done by a force when an object moves a very small distance. Work is a product of force and displacement, and when dealing with variable forces, it helps to break the work into infinitesimally small parts.
The differential work \(dw\) can be calculated using the dot product of the force vector and the infinitesimal displacement vector \(d\vec{r}\). This is given by:
\(dw = \vec{F} \cdot d\vec{r}\).

• This operation considers the force's contribution along the direction of movement.
• When the force and displacement are in the same direction, the work is positive; if opposite, the work is negative.

In this problem, the force depends on position, so calculating \(dw\) involves finding \(dw=(3x\,dx + 4y\,dy)\). This allows us to accumulate work over a path by integrating it later.

Integration is a mathematical process used to find the total value of a quantity when given a rate of change or differential parts. In physics, integration is often used to calculate the total work done by a variable force over a path or period.
To find the total work done by a force vector, integrate the differential work from the starting to the ending position. In this exercise, we are integrating from point (0,0) to point (3,4) in the xy-plane.

• The equation for the total work done is \(W = \int_0^3 3x\,dx + \int_0^4 4y\,dy\).
• The integral of \(3x\,dx\) calculates work along the x-direction, while \(\int 4y\,dy\) accumulates work along the y-direction.

Understanding how to utilize calculus in physics problems is essential for solving real-world problems involving non-constant forces.

Displacement refers to the change in position of an object. It is a vector quantity, which means it has both magnitude and direction.
Displacement is different from distance because it considers the initial and final points only, not the path taken. It is represented as \(d\vec{r}\) in the context of our solution.

• The vector \(d\vec{r}\) can be expressed in its components as \(dx\hat{i} + dy\hat{j}\).
• Calculating displacement helps determine the path of work application by the force on an object in a given direction.

In our exercise, the object moves from the origin \((0,0)\) to the point \((3,4)\), representing a clear displacement that helps determine the complete work done by the force vector.