Problem 43 - University Calculus: Early Transcendentals

Question

Find the work done by a force \(\mathbf{F}=5 \mathbf{i}\) (magnitude \(5 \mathrm{N}\) ) in moving an object along the line from the origin to the point (1,1) (distance in meters).

Step-by-Step Solution

Answer

The work done is 5 Joules.

Step 1: Identify the Force

The force vector is given as \( \mathbf{F} = 5 \mathbf{i} \). This means the force has a magnitude of 5 Newtons in the x-direction.

Step 2: Identify the Path

The object moves along the line from the origin (0,0) to the point (1,1). So the displacement vector of the object will be \( \mathbf{d} = 1 \mathbf{i} + 1 \mathbf{j} \).

Step 3: Calculate the Component of Force along Displacement

To find the component of force along the displacement, we need the dot product of \( \mathbf{F} = 5 \mathbf{i} \) and \( \mathbf{d} = 1 \mathbf{i} + 1 \mathbf{j} \). Calculating the dot product gives:\[ \mathbf{F} \cdot \mathbf{d} = (5 \mathbf{i}) \cdot (1 \mathbf{i} + 1 \mathbf{j}) = 5 \times 1 + 0 \times 1 = 5. \]

Step 4: Conclusion on Work Done

The work done by the force is equal to the dot product calculated in Step 3, which is 5 Joules. Therefore, the work done is 5 Joules.

Key Concepts

Force Vector

A force vector is an arrow that represents both the magnitude and direction of a force. In our example, the force vector is given as \( \mathbf{F} = 5 \mathbf{i} \), which indicates that a force of 5 Newtons is applied in the positive x-direction only. Force vectors are crucial in physics because they allow us to understand how forces act in specific directions.
Some characteristics of a force vector include:

• Magnitude: The size or strength of the force, measured in Newtons. Here, it's 5 N.
• Direction: Indicated by the vector's orientation, in this case, it's entirely in the x-direction.
• Components: A force vector can be broken down into its x, y, and z components. For \( \mathbf{F} = 5 \mathbf{i} \), the y and z-components are zero.

Understanding a force vector's direction and magnitude is vital for calculating the work done by that force along a path.

Dot Product

The dot product is a mathematical operation that combines two vectors into a single scalar value. In terms of physics, it measures how much one vector overlaps with the direction of another. This is especially useful for calculating work because work is defined as the force applied in the direction of displacement times the distance moved.
In the exercise, we calculate the dot product between the force vector, \( \mathbf{F} = 5 \mathbf{i} \), and the displacement vector, \( \mathbf{d} = 1 \mathbf{i} + 1 \mathbf{j} \). The formula for the dot product is:\[ \mathbf{F} \cdot \mathbf{d} = F_x d_x + F_y d_y. \] Applying the values, we find:\[ 5 \cdot 1 + 0 \cdot 1 = 5. \]

• The result, 5, is the amount of force applied in the direction of movement.
• Understanding dot products help us define the component of force that does work.

Thus, the dot product simplifies calculating work when forces and movements aren't aligned.

Displacement Vector

The displacement vector is a mathematical representation of a change in position. In our case, it indicates the object's movement from one point to another. Displacement vectors are expressed in terms of the basic unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) which denote movement in the x, y, and z directions respectively.
For this scenario, the object moves from the origin (0,0) to the point (1,1), giving a displacement vector \( \mathbf{d} = 1 \mathbf{i} + 1 \mathbf{j} \). This tells us that:

• The object moved 1 meter in the x-direction.
• The object moved 1 meter in the y-direction.
• The total path taken can be understood as a straight line in the xy-plane.

Displacement vectors are vital for calculating the work done by a force along a path. They tell us not just how far, but in what specific direction an object is moved, which is essential for applying the dot product correctly.