Question

Calculate the work done in the following situations. A constant force \(\mathbf{F}=\langle 2,4,1\rangle\) (in newtons) moves an object from (0,0,0) to \((2,4,6) .\) (Distance is measured in meters.)

Short answer

Answer: The work done in moving the object is 26 Joules.

Step-by-step solution

Calculate the displacement vector

The displacement vector is the difference between the final position and the initial position of the object. We are given the coordinates of the initial position as (0,0,0) and the final position as (2,4,6). Thus, the displacement vector can be computed as: \(\mathbf{d}=\langle 2-0,4-0,6-0\rangle=\langle 2,4,6\rangle\)

Find the dot product of the force and displacement vectors

To calculate the work done, we need to find the dot product of the force vector \(\mathbf{F}\) and the displacement vector \(\mathbf{d}\). The dot product is given by: \(\mathbf{F} \cdot \mathbf{d} = F_x \cdot d_x + F_y \cdot d_y + F_z \cdot d_z\) where \(F_x,F_y,\) and \(F_z\) represent the x, y, and z components of the force vector, and \(d_x,d_y,\) and \(d_z\) represent the x, y, and z components of the displacement vector. Now substituting the given values we get: \(\mathbf{F} \cdot \mathbf{d} = (2)(2) + (4)(4) + (1)(6) = 4 + 16 + 6 = 26\)

Calculate the work done

Finally, the work done is equal to the dot product of the force vector and the displacement vector. Thus, the work done in moving the object from the initial position to the final position is: Work Done = \(\mathbf{F} \cdot \mathbf{d} = 26 \text{ Joules}\)

Key concepts

Displacement Vector

When an object moves from one position to another, we describe this movement using a displacement vector. This vector is particularly important because it captures the change in position in terms of direction and magnitude.
In our example, the object moves from the initial position

• (0,0,0)

to the final position

• (2,4,6)

The displacement vector is calculated by subtracting the initial position from the final position. As a result, we get:
\[\mathbf{d} = \langle 2-0, 4-0, 6-0 \rangle = \langle 2,4,6 \rangle\]
This vector

• Represents the direction in which the object moved
• Illustrates the distance it covered in the x, y, and z directions

Notice how the displacement vector encapsulates both the direction and the distance. Understanding this concept is crucial when dealing with vector quantities in physics.

Dot Product

The dot product is a key operation when working with vectors, especially in physics. It is used to determine quantities like work, where both direction and magnitude matter. The dot product of two vectors essentially projects one vector onto another and multiplies by the magnitude.
For our case, we calculate the dot product of the force vector

• \(\mathbf{F} = \langle 2,4,1 \rangle\)

and the displacement vector

• \(\mathbf{d} = \langle 2,4,6 \rangle\)

The formula for this is:
\[\mathbf{F} \cdot \mathbf{d} = F_x \cdot d_x + F_y \cdot d_y + F_z \cdot d_z\]
Breaking it down into components, we multiply the corresponding x, y, and z components:

• \(F_x = 2\), \(d_x = 2\)
• \(F_y = 4\), \(d_y = 4\)
• \(F_z = 1\), \(d_z = 6\)

Calculating this gives us:
\[(2)(2) + (4)(4) + (1)(6) = 4 + 16 + 6 = 26\]
Through the dot product, we calculate how much of the force is effectively being used in moving the object along the direction of displacement. This scalar result represents the work done.

Vector Components

Vectors are physical quantities that have both magnitude and direction, and they are often broken down into components to simplify calculations. The components of a vector are essentially the projections of the vector onto the x, y, and z axes.
For a vector \(\mathbf{v} = \langle v_x, v_y, v_z \rangle\), each component represents how much of the vector's magnitude is aligned with each coordinate axis.
In our scenario, the force vector \(\mathbf{F}\) and the displacement vector \(\mathbf{d}\) both have three components:

• Force vector: \(\mathbf{F} = \langle 2, 4, 1 \rangle\)
• Displacement vector: \(\mathbf{d} = \langle 2, 4, 6 \rangle\)

When calculating physical quantities, like work, we often deal with each component separately by using operations like dot product. This way:

• We preserve the directionality of the vectors
• The calculations become manageable and more intuitive

Breaking vectors into components is fundamental for vector addition, scaling, and operations like the dot product, making it a crucial skill in physics.