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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

Expert verified
(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

01

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.
02

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 \( W = 1350 \times 5000 \times \cos(0) = 6,750,000 \, \text{J} \).
03

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, \( W = 1350 \times 5000 \times \cos(35^\circ) \approx 5,532,771 \, \text{J} \).
04

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 \(-5,532,771 \, \text{J}\).
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

An elevator has mass 600 kg, not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 m (five floors) in 16.0 s, and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 kg.

An ingenious bricklayer builds a device for shooting bricks up to the top of the wall where he is working. He places a brick on a vertical compressed spring with force constant \(k = 450\) N/m and negligible mass. When the spring is released, the brick is propelled upward. If the brick has mass 1.80 kg and is to reach a maximum height of 3.6 m above its initial position on the compressed spring, what distance must the bricklayer compress the spring initially? (The brick loses contact with the spring when the spring returns to its uncompressed length. Why?)

Two blocks are connected by a very light string passing over a massless and frictionless pulley (\(\textbf{Fig. E6.7}\)). Traveling at constant speed, the 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward. How much work is done (a) on the 12.0-N block by (i) gravity and (ii) the tension in the string? (b) How much work is done on the 20.0-N block by (i) gravity, (ii) the tension in the string, (iii) friction, and (iv) the normal force? (c) Find the total work done on each block.

A car is traveling on a level road with speed \(v_0\) at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work energy theorem to calculate the minimum stopping distance of the car in terms of \(v_0\), \(g\), and the coefficient of kinetic friction \(\mu_k\) between the tires and the road. (b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h\), making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\). (b) Use the work\(-\)energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if the mass had been dropped from height \(h\), independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom.

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