91Ó°ÊÓ

In physics, the displacement vector is a fundamental concept used to describe how far and in what direction an object has moved from its initial position. Essentially, it's a way to express the change in an object's position.
For the exercise at hand, the object's displacement vector is described as \( (2.0 \, \text{m}) \hat{i} + (-4.0 \, \text{m}) \hat{j} \). This tells us that the object has moved 2 meters along the positive x-axis (\( \hat{i} \) direction) and 4 meters along the negative y-axis (\( \hat{j} \) direction).
This is how displacement differs from distance: while distance measures how much ground an object has covered, displacement gives a detailed vector that includes both length and direction. This vector is critical when calculating the work done on an object, because it includes directional information which is important for finding the dot product with the force vector.

Understanding force components is crucial when dealing with forces acting at an angle. The original exercise involves a force with a known magnitude that makes a particular angle with the x-axis. To effectively use this force in calculations, it must be broken down into components along the x and y axes.
Here's how it's done:

• The x-component (\( F_x \)) of the force can be found using the cosine of the angle: \( F_x = F \cdot \cos(\theta) \). For our exercise, \( F = 10 \, \text{N} \) and \( \theta = 150^\circ \), so \( F_x = 10 \times \left(-\frac{\sqrt{3}}{2}\right) = -5\sqrt{3} \, \text{N} \).
• The y-component (\( F_y \)) is found using the sine of the angle: \( F_y = F \cdot \sin(\theta) \). Hence, \( F_y = 10 \times \frac{1}{2} = 5 \, \text{N} \).

By breaking the force into these components, we can easily apply them in vector mathematics, such as calculating the work done by the force. This approach simplifies the process and ensures accuracy in physics problems.

The dot product, also known as the scalar product, is an important operation in vector algebra. It is particularly useful in physics for calculating work done by a force. In this context, the dot product of two vectors results in a scalar value which represents a physical quantity like work.
For our exercise, the work done by the force is calculated using the dot product of the force vector \( \mathbf{F} \) and the displacement vector \( \mathbf{d} \). This is expressed as:

• \( W = \mathbf{F} \cdot \mathbf{d} = (F_x \cdot d_x) + (F_y \cdot d_y) \)

This formula implies multiplying the corresponding components of the force and displacement vectors, and then summing them up. From the solution provided, we have:

• \( F_x = -5\sqrt{3} \, \text{N} \) and \( d_x = 2.0 \, \text{m} \)
• \( F_y = 5 \, \text{N} \) and \( d_y = -4.0 \, \text{m} \)

Thus, the work done is calculated as:\(W = (-5\sqrt{3})(2.0) + (5)(-4.0) = -10\sqrt{3} - 20 \, \text{Joules}\)
Understanding the dot product helps in grasping how directional forces contribute to work, allowing for the integration of physics and vector mathematics.