Ch 09: Work and Kinetic Energy

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 09: Work and Kinetic Energy

Problem 18

Chapter 9, Problem 18

The three ropes shown in the bird's-eye view of FIGURE EX9.18 are used to drag a crate 3.0 m across the floor. How much work is done by each of the three forces?

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Identify the forces acting on the crate. From the problem, there are three forces applied via ropes. Let these forces be $ F_1 $, $ F_2 $, and $ F_3 $, each with a specific magnitude and direction. The directions of the forces are given in the figure (not provided here). Work is calculated for each force individually.

Recall the formula for work: $ W = F \cdot d \cdot \cos(\theta) $, where $ F $ is the magnitude of the force, $ d $ is the displacement (3.0 m in this case), and $ \theta $ is the angle between the force and the direction of displacement.

For each force, determine the angle $ \theta $ between the force vector and the displacement vector. Use the information provided in the figure to identify these angles.

Substitute the values of $ F $, $ d $, and $ \cos(\theta) $ into the work formula for each force. For example, for $ F_1 $, the work done is $ W_1 = F_1 \cdot 3.0 \cdot \cos(\theta_1) $. Repeat this process for $ F_2 $ and $ F_3 $.

Add the calculated work values for each force if the problem requires the total work done on the crate. Otherwise, report the work done by each force individually as $ W_1 $, $ W_2 $, and $ W_3 $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Work

In physics, work is defined as the product of the force applied to an object and the distance over which that force is applied, specifically in the direction of the force. Mathematically, it is expressed as W = F × d × cos(θ), where W is work, F is the force, d is the distance, and θ is the angle between the force and the direction of motion. Understanding this concept is crucial for calculating the work done by each rope in moving the crate.

Force Components

When multiple forces act on an object, it is essential to analyze their components, particularly in relation to the direction of motion. Forces can be broken down into horizontal and vertical components, which can be calculated using trigonometric functions. This breakdown allows for a clearer understanding of how much of each force contributes to the work done in moving the crate across the floor.

Friction

Friction is the resistive force that opposes the motion of an object sliding across a surface. It is influenced by the nature of the surfaces in contact and the normal force acting on the object. In the context of dragging the crate, understanding friction is vital, as it affects the net force acting on the crate and, consequently, the total work done by the applied forces.

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Textbook Question

FIGURE EX9.20 is the force-versus-position graph for a particle moving along the x-axis. Determine the work done on the particle during each of the three intervals 0–1 m, 1–2 m, and 2–3 m.

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Textbook Question

You throw a 5.5 g coin straight down at 4.0 m/s from a 35-m-high bridge. What is the speed of the coin just as it hits the water?

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Textbook Question

A particle moving on the x-axis experiences a force given by F_x = qx², where q is a constant. How much work is done on the particle as it moves from x = 0 to x = d?

A 25 kg air compressor is dragged up a rough incline from ${\vec{r}}_{1} = (1.3 \hat{ı} + 1.3 \hat{ȷ}) m$ to ${\vec{r}}_{2} = (8.3 \hat{ı} + 2.9 \hat{ȷ}) m$ , to where the y-axis is vertical. How much work does gravity do on the compressor during this displacement?

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Textbook Question

A 45 g bug is hovering in the air. A gust of wind exerts a force $\vec{F} = (4.0 \hat{ı} - 6.0 \hat{ȷ}) \times 10^{- 2} N$ on the bug. What is the bug's speed at the end of this displacement? Assume that the speed is due entirely to the wind.

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Textbook Question

A 2.0 kg particle moving along the x-axis experiences the force shown in FIGURE EX9.22. The particle's velocity is 3.0 m/s at x = 0 m. At what point on the x-axis does the particle have a turning point?

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