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 situations, where \(x\) is in meters: $$\begin{array}{ll} \text { Outward } & \text { Inward } \\ \hline \text { (a) }+3.0 & -3.0 \\ \text { (b) }+5.0 & +5.0 \\ \text { (c) }+2.0 x & -2.0 x \\ \text { (d) }+3.0 x^{2} & +3.0 x^{2} \\ \hline \end{array}$$ 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

Understand Work Done

To find the net work done on the particle, we need to calculate the work done during the outward trip and the return trip, and then sum them up. Work done, \( W \), by the force during a movement from position \( x_1 \) to \( x_2 \) is given by \( W = \int_{x_1}^{x_2} F(x) \ dx \), where \( F(x) \) is the force function.

02

Calculate Work for Situation (a)

For situation (a), the force is constant: \( F(x) = +3.0 \) N for the outward trip and \( F(x) = -3.0 \) N for the return trip.For the outward trip: \[ W_{ ext{outward}} = \int_{1}^{4} 3 \, dx = 3 \times (4 - 1) = 9 \text{ J} \]For the return trip: \[ W_{ ext{return}} = \int_{4}^{1} -3 \, dx = -3 \times (1 - 4) = 9 \text{ J} \]Net work done: \[ W_{ ext{net}} = 9 + 9 = 18 \text{ J} \]

03

Calculate Work for Situation (b)

For situation (b), the force is constant: \( F(x) = +5.0 \) N for both the outward and the return trips.For the outward trip: \[ W_{ ext{outward}} = \int_{1}^{4} 5 \, dx = 5 \times (4 - 1) = 15 \text{ J} \]For the return trip: \[ W_{ ext{return}} = \int_{4}^{1} 5 \, dx = 5 \times (1 - 4) = -15 \text{ J} \]Net work done: \[ W_{ ext{net}} = 15 + (-15) = 0 \text{ J} \]

04

Calculate Work for Situation (c)

For situation (c), \( F(x) = +2.0x \) for the outward trip and \( F(x) = -2.0x \) for the return trip.For the outward trip:\[ W_{ ext{outward}} = \int_{1}^{4} 2x \, dx = \left[ x^2 \right]_{1}^{4} = 4^2 - 1^2 = 23 \text{ J} \]For the return trip:\[ W_{ ext{return}} = \int_{4}^{1} -2x \, dx = -\left[ x^2 \right]_{4}^{1} = -(1^2 - 4^2) = -23 \text{ J} \]Net work done: \[ W_{ ext{net}} = 23 + (-23) = 0 \text{ J} \]

05

Calculate Work for Situation (d)

For situation (d), \( F(x) = +3.0x^2 \) for both the outward and the return trips.For the outward trip:\[ W_{ ext{outward}} = \int_{1}^{4} 3x^2 \, dx = \left[ x^3 \right]_{1}^{4} = \frac{3}{3}(4^3 - 1^3) = 63 \text{ J} \]For the return trip:\[ W_{ ext{return}} = \int_{4}^{1} 3x^2 \, dx = \frac{3}{3}(1^3 - 4^3) = -63 \text{ J} \]Net work done: \[ W_{ ext{net}} = 63 + (-63) = 0 \text{ J} \]

06

Determine Conservative Force

A force is conservative if the net work done by it in a round trip is zero. Based on our calculations: - Situation (a) has a net work of 18 J, so the force is not conservative. - Situations (b), (c), and (d) all have a net work of 0 J, so the forces in these situations are conservative.

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Key Concepts

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

Net Work Calculations

Calculating net work involves understanding how much work the force performs over an entire path. To find the net work, combine the work done over separate parts of the path. In our exercise, we have two trips: outward and return. Each trip can involve different forces, which should be calculated separately.

If the force is constant, you can use a simpler calculation:

• Multiply the force by the distance covered.
• Take the direction into account by using a positive or negative sign.

For example, if you move from position \( x_1 \) to \( x_2 \) under a constant force, calculate it as \( F \times (x_2 - x_1) \). Don't forget that each leg of the journey might have different force values.

The key takeaway is summing the work calculated for each leg of the journey. If net work is zero, it means total energy is conserved over the round trip.

Conservative Forces

A conservative force is a type of force where the total work done around a closed loop is zero. This means no net energy is added to or taken from the system in a complete cycle. Recognizing whether forces in a problem are conservative helps us understand energy conservation in the system.

What indicates a conservative force?

• Net work done over any closed path is zero.
• Examples include gravity and spring forces.

In our exercise, when a force results in zero net work after moving back and forth, it is conservative. This suggests that energy is conserved entirely and no external force is performing work on the system over the cycle.

Integral Calculations

Integral calculations help us find the work done by a variable force. If the force changes with position, a simple multiplication with distance doesn't suffice. Instead, integrals allow us to calculate work by summing infinitesimal contributions over the path.

Here's how to approach it:

• The integral of the force function \( F(x) \) over a path \( x_1 \) to \( x_2 \) calculates total work.
• The formula is \( W = \int_{x_1}^{x_2} F(x) \, dx \).

In our situations with forces like \( 2.0x \) and \( 3.0x^2 \), these are not constant. Hence, the integral \( \int (F(x)) \, dx \) should be calculated over the given limits to accurately determine work. This process ensures every bit of force along the path's contribution is accounted for, leading to precise calculations of work done.