Question

Using a cable with a tension of 1350 N, a tow truck pulls a car 5.00 km along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at 35.0\(^\circ\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

Short answer

(a) 6,750,000 J and 5,532,771 J. (b) -6,750,000 J and -5,532,771 J. (c) 0 J.

Step-by-step solution

Understand the Work Formula

Work done by a force is calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( W \) is the work done, \( F \) is the force applied, \( d \) is the distance moved in the direction of the force, and \( \theta \) is the angle between the force and the direction of motion.

Calculate Work for Horizontal Pull

For a horizontal pull, the angle \( \theta \) is 0 degrees. Substitute into the work formula: \( F = 1350 \, \text{N} \), \( d = 5000 \, \text{m} \), and \( \theta = 0^\circ \). Thus, the work done is \( W = 1350 \times 5000 \times \cos(0) = 6,750,000 \, \text{J} \).

Calculate Work for Pull at 35 Degrees

For an angle \( \theta = 35^\circ \), substitute into the work formula: \( F = 1350 \, \text{N} \), \( d = 5000 \, \text{m} \), and \( \theta = 35^\circ \). Thus, \( W = 1350 \times 5000 \times \cos(35^\circ) \approx 5,532,771 \, \text{J} \).

Evaluate Work Done on Tow Truck

In both scenarios from part (a), the work done by the cable on the tow truck is equal and opposite to the work done on the car. Thus, when the force is horizontal, it is \(-6,750,000 \, \text{J}\) and when the force is at 35 degrees, it is \(-5,532,771 \, \text{J}\).

Assess Work Done by Gravity

Since the pull is horizontal and there is no vertical displacement, the work done by gravity on the car for both scenarios in part (a) is zero because gravity acts perpendicular to the direction of motion.

Key concepts

Work Done by a Force

The concept of work in physics involves how a force causes an object to move over a distance. To calculate work done by a force, we use the formula:

• \( W = F \cdot d \cdot \cos(\theta) \)

Here, \( W \) represents work, \( F \) is the force applied, \( d \) is the distance covered, and \( \cos(\theta) \) is the cosine of the angle \( \theta \) between the direction of the force and the direction of movement.
To understand how much work is done, think of the force as needing a specific path. If the force doesn't contribute to the distance in its direction, no work is recorded along that axis. Thus, the angle \( \theta \) plays a key role in determining how effectively the force contributes to performing the work.

Angle of Force Application

The angle at which a force is applied significantly affects the amount of work done. When a force is applied in the exact direction of motion, the angle \( \theta \) is 0 degrees, and the full magnitude of the force contributes to the work done. This is why \( \cos(0) = 1 \).
As the angle increases away from this direct line, the cosine value decreases, indicating that only a component of the force is doing work in the direction of movement.
For example, when the angle is 35 degrees, the cosine function reduces the effective force, making it less impactful than when fully aligned. Therefore, understanding the angle helps predict how much force contributes to moving an object in a particular direction.

Horizontal and Angled Forces

Forces acting horizontally and at an angle have different outcomes in their work applications. A horizontally applied force at 0 degrees has the maximum effect because it aligns perfectly with the movement direction, thus converting all its energy into moving the object.
Angled forces, like one at 35 degrees, have to be broken down into components. The calculation involves using trigonometry to find how much of that force moves the object horizontally and how much might contribute to other effects, like lifting the object somewhat.

• Horizontal forces: Full force acts on movement.
• Angled forces: Effectiveness depends on cosine of the angle.

Though the applied force remains the same, its horizontal component decreases with more inclination, affecting the total work done on the object.

Gravitational Work

Gravitational work focuses on the direction and impact of gravity on moving objects. Gravity acts vertically downward. Thus, if there is no vertical movement in an object's motion, gravity doesn't perform work on it.
In scenarios like a tow truck pulling a car horizontally, gravity's force is perpendicular to the horizontal motion. This alignment means gravity does no work along the car's path because its displacement occurs without any vertical distance.
Without vertical displacement, gravitational work remains zero, even if the object weighs significantly. This concept simplifies calculations where vertical contributions by forces like gravity are not part of the analysis.

Physics Problem Solving

Problem-solving in physics often involves breaking down complex scenarios into manageable calculations. The fundamental relationship between force, distance, and angle helps simplify real-world mechanics into calculable work.
Here's a simple step-by-step approach to tackle such problems:

• Identify the forces in action and their directions.
• Determine the distance over which these forces are playing a role.
• Assess the angle of each force with respect to direction of movement.

Applying these steps to solve problems provides a clearer picture of material forces and interactions. By understanding each aspect separately, students can solve complex physics scenarios systematically, ensuring accuracy and the comprehensive application of formulas.