Problem 14 - University Physics with Modern Physics

Question

You apply a constant force \(\overrightarrow{F} =(-68.0 \, \mathrm{N})\hat{\imath} +(36.0 \, \mathrm{N})\hat{\jmath}\) to a 380-kg car as the car travels 48.0 m in a direction that is 240.0\(^\circ\) counterclockwise from the +\(x\)-axis. How much work does the force you apply do on the car?

Step-by-Step Solution

Answer

The work done on the car is 135.552 J.

Step 1: Understand the concept of work

The work done (W) by a force is given by the dot product of the force vector \( \overrightarrow{F} \) and the displacement vector \( \overrightarrow{s} \). The formula for work is: \[ W = \overrightarrow{F} \cdot \overrightarrow{s} = F_x s_x + F_y s_y \] where \( F_x \) and \( F_y \) are the components of the force, and \( s_x \) and \( s_y \) are the components of the displacement.

Step 2: Calculate displacement components

The car travels in a direction of 240.0° counterclockwise from the +x-axis. Break the displacement vector \( \overrightarrow{s} \) into its x and y components. The displacement is 48.0 m long. Use trigonometric functions: \[ s_x = 48.0 \cos(240.0^{\circ}) \]\[ s_y = 48.0 \sin(240.0^{\circ}) \] Calculating gives: \( s_x = 48.0 (-0.5) = -24.0 \, \text{m} \) and \( s_y = 48.0 (-0.866) = -41.568 \, \text{m} \).

Step 3: Use the work formula

Substitute the components of force and displacement into the work formula: \[ W = (-68.0 \, \text{N})(-24.0 \, \text{m}) + (36.0 \, \text{N})(-41.568 \, \text{m}) \] Calculating gives: \( W = 1632.0 \, \text{N}\cdot\text{m} - 1496.448 \, \text{N}\cdot\text{m} = 135.552 \, \text{J} \).

Step 4: State the amount of work done

Thus, the work done by the force \( \overrightarrow{F} \) on the car as it travels 48.0 m is 135.552 J.

Key Concepts

Force Vector

A force vector is a way to express the magnitude and direction of a force applied to an object. In the exercise, the force vector \( \overrightarrow{F} \) is given as \((-68.0 \, \mathrm{N})\hat{\imath} + (36.0 \, \mathrm{N})\hat{\jmath}\). This notation is called unit vector notation, where:

• \(\hat{\imath}\) represents the component in the x-direction.
• \(\hat{\jmath}\) represents the component in the y-direction.

Each of these components shows how much force is applied along the respective axis. The negative sign for the x-component indicates the direction of the force is opposite to the positive x-direction. Understanding force vectors is crucial in determining how forces affect the movement of objects in physics.

Displacement Vector

The displacement vector represents how far and in what direction an object moves from its initial position. In the context of this exercise, the car's movement is described by the displacement vector \( \overrightarrow{s} \). The car travels a distance of 48.0 m in a direction specified by an angle of 240.0° counterclockwise from the positive x-axis.
The displacement vector can be broken into two components, using trigonometric functions:

• The x-component \( s_x = 48.0 \cos(240.0^{\circ}) \)
• The y-component \( s_y = 48.0 \sin(240.0^{\circ}) \)

These calculations use the cosine and sine of the given angle to find the respective components, showing how the total displacement divides between horizontal and vertical directions.

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually vectors) and returns a single number. In physics, this is used to calculate work done by a force, as work is the dot product of the force vector \( \overrightarrow{F} \) and the displacement vector \( \overrightarrow{s} \). The formula for work is:
\[ W = \overrightarrow{F} \cdot \overrightarrow{s} = F_x s_x + F_y s_y \]

• \( F_x \) and \( F_y \) are the components of the force.
• \( s_x \) and \( s_y \) are the components of the displacement.

The dot product can tell us how much of the force is used in doing work in the direction of the displacement. It's important in calculating realistic scenarios where forces do not act in the same direction as movement.

Trigonometry in Physics

Trigonometry is a branch of mathematics that studies the relations between side lengths and angles of triangles. In physics, it is often used to resolve vectors, like forces and displacements, into their components.
In this exercise, to find the components of the displacement vector \( \overrightarrow{s} \), we utilize basic trigonometric functions:

• \(\cos\) to find the horizontal component \( s_x \)
• \(\sin\) to find the vertical component \( s_y \)

The angle provided (240.0 degrees) is used in conjunction with these functions to determine how the 48.0 m displacement splits into horizontal and vertical parts. Understanding how to apply trigonometry in this context is crucial for solving many problems involving vectors in physics.