Question

Find the work done by a force of magnitude 10 newtons acting in the direction of the vector \(3 \boldsymbol{i}+\boldsymbol{j}+8 \boldsymbol{k}\) if it moves a particle from the point \((1,1,1)\) to the point \((3,1,2)\).

Short answer

Answer: The work done by the force on the particle is \(W = \frac{140}{\sqrt{74}} \text{ J}\).

Step-by-step solution

Find the displacement vector

The displacement vector can be found by subtracting the initial position of the particle from its final position. Let the initial position vector be \(\boldsymbol{A} = \boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k}\) and the final position vector be \(\boldsymbol{B} = 3\boldsymbol{i} + \boldsymbol{j} + 2\boldsymbol{k}\). Then, the displacement vector \(\boldsymbol{D}\) is: $$ \boldsymbol{D} = \boldsymbol{B} - \boldsymbol{A} = (3\boldsymbol{i} + \boldsymbol{j} + 2\boldsymbol{k}) - (\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k}) $$

Simplify the displacement vector

Simplifying the equation for the displacement vector, we get: $$ \boldsymbol{D} = 2\boldsymbol{i} + 0\boldsymbol{j} + 1\boldsymbol{k} $$

Normalize the given force vector

Since the magnitude of the given force vector is 10 N, we need to normalize the force vector which has components as \(3\boldsymbol{i}+\boldsymbol{j}+8\boldsymbol{k}\). The normalization formula is \(\frac{\boldsymbol{F}}{\lVert \boldsymbol{F} \rVert}\), where \(\lVert \boldsymbol{F} \rVert\) represents the magnitude of the force vector. First, find the magnitude of the force vector: $$ \lVert \boldsymbol{F} \rVert = \sqrt{(3)^2 + (1)^2 + (8)^2} = \sqrt{74} $$ Next, normalize the force vector: $$ \frac{\boldsymbol{F}}{\lVert\boldsymbol{F}\rVert} = \frac{3\boldsymbol{i} + \boldsymbol{j} + 8\boldsymbol{k}}{\sqrt{74}} $$

Multiply the normalized force vector by its magnitude

Now, we multiply the normalized force vector by the given magnitude of 10 N to find the force vector with the correct magnitude: $$ \boldsymbol{F} = 10 \cdot \frac{3\boldsymbol{i} + \boldsymbol{j} + 8\boldsymbol{k}}{\sqrt{74}} $$

Calculate the dot product of the force vector and the displacement vector

Finally, we calculate the dot product of the force vector and the displacement vector to find the work done: $$ W = \boldsymbol{F} \cdot \boldsymbol{D} = 10 \cdot \frac{3\boldsymbol{i} + \boldsymbol{j} + 8\boldsymbol{k}}{\sqrt{74}} \cdot (2\boldsymbol{i} + 0\boldsymbol{j} + 1\boldsymbol{k}) $$ $$ W = 10 \cdot \frac{(3\cdot 2) + (1\cdot 0) + (8\cdot 1)}{\sqrt{74}} = 10 \cdot \frac{14}{\sqrt{74}} $$

Simplify the expression for work done

The work done by the force is: $$ W = \frac{140}{\sqrt{74}} \text{ J} $$

Key concepts

Vector Arithmetic

In physics and mathematics, vector arithmetic is a fundamental tool used to describe and manipulate quantities that have direction and magnitude, such as forces or velocities.
To understand how vector arithmetic applies to calculating work, let's start by considering basic operations involving vectors: addition, subtraction, and scalar multiplication.
When a particle moves from one point to another, its change in position is represented by a displacement vector, which can be found by subtracting the initial position vector from the final position vector. In the context of the given exercise, if position A is \((1,1,1)\) and position B is \((3,1,2)\), the displacement vector \(\boldsymbol{D}\) is calculated using vector subtraction:
\[\boldsymbol{D} = \boldsymbol{B} - \boldsymbol{A}\]
This vector arithmetic operation gives us the direction and length of the straight path from A to B, which is crucial for finding the work done by a force along this path.

Dot Product

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is vital when calculating work done by a force because work is defined as the dot product of the force vector and the displacement vector.
In the general form, the dot product of two vectors \(\boldsymbol{A}\) and \(\boldsymbol{B}\) is given by:
\[\boldsymbol{A} \. \boldsymbol{B} = \sum \boldsymbol{A}_i \times \boldsymbol{B}_i\]
Here, \(\boldsymbol{A}_i\) and \(\boldsymbol{B}_i\) represent the components of vectors \(A\) and \(B\), respectively. For our problem, once we have the force vector and displacement vector, their dot product gives us the work done (W), as it is the product of the component of the force in the direction of the displacement and the magnitude of the displacement.

Force Vector Normalization

Force vector normalization is a method used to simplify calculations in physics, particularly when dealing with vectors representing force.
Normalization refers to the process of converting an arbitrary vector to a unit vector (a vector with a magnitude of 1) that points in the same direction. This is done by dividing the vector by its magnitude. The formula for a normalized vector \(\hat{\boldsymbol{F}}\) is:
\[\hat{\boldsymbol{F}} = \frac{\boldsymbol{F}}{\lVert \boldsymbol{F} \rVert}\]
In the exercise, we first calculate the magnitude \(\lVert \boldsymbol{F} \rVert\) of the force vector, then use it to normalize the vector. Once normalized, we can easily scale the unit vector to any desired magnitude—in this case, the magnitude of the force (10 N). Normalizing the force vector is essential for the accurate computation of work done as it ensures that the force vector's direction is properly accounted for.

Displacement Vector

A displacement vector represents the shortest distance from the initial to the final position of a point and is direction-aware. In physical terms, it denotes the change in position of an object. It is often symbolized with an arrow, with its length proportional to the displacement's magnitude and its direction from start to end point.
In our exercise, the displacement vector is critical for determining the work done. It is found using vector arithmetic and then used in the dot product with the force vector. The displacement vector \(\boldsymbol{D}\) for a particle moving from point \((1,1,1)\) to \((3,1,2)\) is simplified to \(\boldsymbol{D} = 2\boldsymbol{i} + 0\boldsymbol{j} + 1\boldsymbol{k}\), revealing that the particle moved 2 units in the x-direction and 1 unit in the z-direction, with no movement in y-direction. This vector not only shows how far the particle moved, but also in which direction, both needed to calculate work done.