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Ch 09: Work and Kinetic Energy
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 09: Work and Kinetic EnergyProblema 20
Capitolo 9, Problema 20

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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Step 1: Understand the concept of work done. Work done (W) is calculated as the area under the force (F) versus position (x) graph. For each interval, calculate the area under the graph corresponding to that interval.
Step 2: Convert the x-axis units from cm to m, as the problem specifies intervals in meters. For example, 0–1 m corresponds to 0–100 cm, 1–2 m corresponds to 100–200 cm, and 2–3 m corresponds to 200–300 cm.
Step 3: For the interval 0–1 m (0–100 cm), observe that the force is constant at 2 N. The area under the graph is a rectangle with width 1 m and height 2 N. Use the formula for the area of a rectangle: \( \text{Area} = \text{Force} \times \text{Displacement} \).
Step 4: For the interval 1–2 m (100–200 cm), observe that the force increases linearly from 2 N to 6 N. The area under the graph is a trapezoid. Use the formula for the area of a trapezoid: \( \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \), where the bases are the force values and the height is the displacement.
Step 5: For the interval 2–3 m (200–300 cm), observe that the force decreases linearly from 6 N to 0 N. The area under the graph is a triangle. Use the formula for the area of a triangle: \( \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \), where the base is the displacement and the height is the force.

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Work Done by a Force

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 θ is the angle between the force and the direction of motion. In the context of a force-versus-position graph, the work done can be calculated as the area under the curve of the graph for the specified intervals.
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Work Done by a Constant Force

Area Under the Force-Position Graph

The area under a force versus position graph represents the work done on the particle as it moves through a given distance. For linear segments, this area can be calculated using geometric shapes such as rectangles and triangles. The total work done over an interval can be found by summing the areas of these shapes, which correspond to the force applied over the distance traveled.
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Calculating Work As Area Under F-x Graphs

Intervals of Motion

In this problem, the motion of the particle is divided into three distinct intervals: 0–1 m, 1–2 m, and 2–3 m. Each interval may have different forces acting on the particle, which affects the work done in each segment. Understanding the behavior of the force in each interval is crucial for accurately calculating the total work done, as the force may vary in magnitude and direction across these segments.
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