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A constant force is a force that remains the same in both magnitude and direction, irrespective of time or the position of the object it acts upon. When dealing with constant forces, the calculation of work becomes much simpler because you do not have to worry about fluctuations in force magnitude or direction over the course of an object's movement.
In the given exercise, the force is defined as \( \vec{F} = 3 \hat{i} + 4 \hat{j} \). This means the force has a constant component of 3 units in the x-direction and 4 units in the y-direction.
Certainly, a key aspect of constant force is that the work done by such a force depends only on the start and endpoints of the displacement, not on the path taken.
This is crucial when calculating work done over different paths between the same two points, as in our exercise.

A displacement vector describes the change in position of an object. It points from the start position to the end position and carries both magnitude and direction. In the context of work done by a force, the displacement vector indicates the movement over which the force acts.
For the first path in the exercise, the displacement from \((0,0)\) to \((a,a)\) results in a vector \( \vec{r}_1 = (a, a) \). This simply represents moving a units in both the x and y directions simultaneously.
For the second path, the movement is along two segments. First, a displacement to the right along the x-axis \((a, 0)\), followed by movement upward along the y-axis \((0, a)\).
Understanding the components of the displacement vector is vital as it guides how you integrate the force acting over these components to compute work done.

The dot product, also known as the scalar product, is a fundamental operation in vector calculus that combines two vectors and returns a scalar. In simple terms, it multiplies corresponding components of the vectors and sums them up.
In work calculation, the dot product allows us to determine how much of the force acts in the direction of the displacement, thereby quantifying the work done. The formula used is: \( \vec{F} \cdot \vec{r} = F_x \cdot r_x + F_y \cdot r_y \).
This means that only the components of force parallel to the displacement contribute to work done. In Path 1 of the problem, the computation was straightforward using the single displacement vector \((a,a)\). For Path 2, work was calculated along each segment (\(OQ\) and \(QP\)), showing that even though the path differed, combining segment works replicated the net displacement work seen in Path 1.
This solidifies the idea that for work done by a constant force, it's all about the endpoints of the displacement.