91Ó°ÊÓ

We move a particle along an \(x\) axis, first outward from \(x=1.0 \mathrm{~m}\) to \(x=4.0 \mathrm{~m}\) and then back to \(x=1.0 \mathrm{~m},\) while an external force acts on it. That force is directed along the \(x\) axis, and its \(x\) component can have different values for the outward trip and for the return trip. Here are the values (in newtons) for four situations, where \(x\) is in meters: $$ \begin{array}{ll} \hline \text { Outward } & \text { Inward } \\ \hline \text { (a) }+3.0 & -3.0 \\ \text { (b) }+5.0 & +5.0 \\ \text { (c) }+2.0 x & -2.0 x \\ \text { (d) }+3.0 x^{2} & +3.0 x^{2} \\ \hline \end{array} $$ Find the net work done on the particle by the external force for the round trip for each of the four situations. (e) For which, if any, is the external force conservative?

Step by step solution

01

Understanding Net Work (a)

To find the net work done on the particle for the round trip in situation (a), we note that the force remains constant. For the outward trip, the work done is the force times displacement: \[ W_{outward} = F \cdot \Delta x = 3.0 \times (4.0 - 1.0) = 9.0 \, \text{J} \]. For the return trip, the force direction is opposite to the displacement, giving: \[ W_{inward} = -3.0 \times (1.0 - 4.0) = 9.0 \, \text{J} \]. The net work done over the round trip is:\[ W_{net} = W_{outward} + W_{inward} = 9.0 + 9.0 = 18.0 \, \text{J} \].

02

Understanding Net Work (b)

In this situation, the force remains constant but is the same in both directions. For the outward trip, the work is:\[ W_{outward} = 5.0 \times (4.0 - 1.0) = 15.0 \, \text{J} \].For the return trip:\[ W_{inward} = 5.0 \times (1.0 - 4.0) = -15.0 \, \text{J} \].The net work done over the round trip is:\[ W_{net} = W_{outward} + W_{inward} = 15.0 - 15.0 = 0.0 \, \text{J} \].

03

Understanding Net Work (c)

Here, the force depends on position. For the outward trip, we integrate the force function to find work:\[ W_{outward} = \int_{1}^{4} 2.0x \, dx = 2.0 \cdot \left[ \frac{x^2}{2} \right]_{1}^{4} = 2.0 \cdot \left(16 - 1\right) = 30.0 \, \text{J} \].For the return trip:\[ W_{inward} = \int_{4}^{1} -2.0x \, dx = -2.0 \cdot \left[ \frac{x^2}{2} \right]_{4}^{1} = -2.0 \cdot \left(1 - 16\right) = 30.0 \, \text{J} \].Net work over the round trip is:\[ W_{net} = 30.0 + 30.0 = 60.0 \, \text{J} \].

04

Understanding Net Work (d)

The force depends on the square of position, so we integrate:\[ W_{outward} = \int_{1}^{4} 3.0x^2 \, dx = 3.0 \cdot \left[ \frac{x^3}{3} \right]_{1}^{4} = 3.0 \cdot \left(64 - 1\right) = 189.0 \, \text{J} \].For the backward trip:\[ W_{inward} = \int_{4}^{1} 3.0x^2 \, dx = 3.0 \cdot \left[ \frac{x^3}{3} \right]_{4}^{1} = -3.0 \cdot \left(1 - 64\right) = -189.0 \, \text{J} \].Net work is:\[ W_{net} = 189.0 - 189.0 = 0.0 \, \text{J} \].

05

Identifying Conservative Forces

A force is conservative if the net work done over a closed path is zero. From our calculations, situations (b) and (d) resulted in a net work of zero. Thus, the forces in these scenarios are conservative.

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Key Concepts

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

Conservative Forces

Conservative forces are a fundamental concept in physics. These are forces for which the work done is independent of the path taken but depends only on the initial and final positions. This unique property allows energy to be conserved, simplifying the calculations of potential energy.
For a conservative force:

• The net work done around a closed path is zero. This means if you return to your starting point, the work done by the force is zero.
• Examples include gravitational forces, elastic spring forces, and electrostatic forces between charged particles.

In the problem scenario, scenarios (b) and (d) identified zero net work over the round trips, indicating that the forces involved are conservative. Being path-independent, these forces allow for energy conservation in these specific cases.

Net Work

Net work involves calculating the total work done on an object over a path, considering both direction and magnitude of force. It is the cumulative effect of all forces acting on an object when moved between two points.
Net work formula is given by:

• Net work: \( W_{net} = W_{outward} + W_{inward} \)

In the exercises, positive work represents work done to move the particle in the same direction as the force, while negative work is done against it. For example, in scenario (b), where the force remains constant at \( +5.0 \) N for both outward and inward paths, net work was zero since work done on outward and return trips canceled out each other.

Integration of Force

When the force varies with position, calculating work requires integration. Integration allows us to account for continuous changes in force as the position changes.
To calculate work through integration:

• Identify the force function, which can depend on position, such as \( F(x) = 2.0x \).
• Perform the integral over the specified path, \( \int_{x_1}^{x_2} F(x) \, dx \).

In the problem, scenario (c) showcased integration, where \( F = 2.0x \) resulted in a work calculation through solving the integral, showing how work is calculated when force varies with position. Understanding this process enhances comprehension of how work is distributed along a path under varying forces.