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Positive vs. Negative Work

by no_audio
Tags: negative, positive, work
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no_audio
#1
Feb26-10, 08:29 PM
P: 4
What is the difference between positive work and negative work? Because the definition of work is the dot product of the force and displacement vectors, I can see that work is negative when the force acts in the "opposite" direction of the displacement. Also, I see that W = -(delta)PE. I ask this question as a result of encountering two concepts that seem to contradict these statements. The first:

(1) "In a system of two positively charged particles, PE is positive, and positive work must be done to bring like charges together. In a system of two charges opposite in sign, PE is negative, and negative work must be done to bring opposite charges together. Energy is released!"

The second:

(2) "Oppositely charged parallel plates are separated by 5.33 mm. A potential difference of 600 V exists between the plates. How much work must be done on an electron to move it to the negative plate if it is initially positioned 2.90 mm from the positive plate?"





In (1), I can see that the PE is positive when the charges are positive, but shouldn't the work necessary to bring the charges together be negative (since the charges would naturally repel, it's intuitive to me to think that since the displacement works in a direction opposite the electric field force, the work is negative; also, W=-(delta)PE)?

In (2), the answer is 4.38 x 10^-17 J. However, I do not understand why it is a positive answer. Shouldn't the work be negative if the electron is being displaced against the electric field force?



I have a feeling that my questions stem from a misunderstanding of the concept of work. I'm sorry I didn't use the template; I didn't feel that it applied to my questions.
My first post on this forum, yay! I'm so glad I found this site. I'm trying to self-study for the AP Physics B test and I feel this site will help clear up a lot of confusion.
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collinsmark
#2
Feb26-10, 10:00 PM
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Hello no_audio,

Welcome to Physics Forums!

I agree that determining what is positive and what is negative can be confusing. One reason is because in a closed, conservative system, whenever something is losing energy, there is guaranteed to be something else that is gaining energy. And in a conservative open system (where somebody or something is doing work, but the details are not known or given) it can get confusing.

Let me give you an analogy/tool that I like to use. Put yourself in the position of whatever it is that is actually doing the work, whether its specified in the problem or not. You are the one putting the effort into whatever it is that is being done. If you are pushing something, then you are putting energy into it (ignoring friction here - we are speaking of conservative forces). If you are pushing a block forward, accelerating it forward (in the same direction as its velocity), it is increasing in kinetic energy. You are doing positive work on it. Assume you are accelerating a block forward on the positive x axis, and its velocity is also positive x.

[tex] W = \vec F_x \cdot \vec{s}_x [/tex]

Since the velocity is moving along the positive x axis, it implies that from the time you start till the time you finish, the block has moved in position along the positive x axis. That's why [tex] s_x [/tex] is positive. The end result is positive. You have done positive work on the block, and it has gained energy as a result. The sytem has gained energy (i.e. the change in the system's energy is positive)

But you, yourself, have actually lost energy in the process. Consider rearranging the equation a little and you'll see

[tex] 0 = -W + \vec F_x \cdot \vec{s}_x [/tex]

That -W corresponds to your increase in energy, which is actually a decrease in this example. But as it works out, most people are only interested in the system, so your net energy change is not discussed.

Okay, let's make a change, and move to a different example. Suppose we have the same situation, but the velocity in the block is in the opposite direction. You are still pushing on the block in the same direction, but since it's moving toward you, you're slowing it down this time.

In this case we can assume that [tex] s_x [/tex] is negative. Because from start to finish, it's moved in the negative x direction. So now we have,

[tex] W = \vec F_x \cdot (-\vec{s}_x) [/tex]

which is,

[tex] W = -\vec F_x \cdot \vec{s}_x [/tex]

In this case, the block is losing energy. It's doing the work on you! And by the same respect, you are actually gaining energy on this one (assuming you are the one doing the pushing along the positive x axis). So in this example, the work that you do on the block is negative. The system's energy decreases (i.e. its change in energy is negative).

Quote Quote by no_audio View Post
In (1), I can see that the PE is positive when the charges are positive, but shouldn't the work necessary to bring the charges together be negative (since the charges would naturally repel, it's intuitive to me to think that since the displacement works in a direction opposite the electric field force, the work is negative; also, W=-(delta)PE)?
Okay, here is one of those conservative, open systems, where it doesn't really specify who or what is doing the work (moving the charge in this case). So put yourself in there! Now you are the one who is going to push that charge in from infinity.

Imagine yourself pushing this positive charge toward another positive charge. You are the one putting work and effort into this. So you are giving energy to the charge which you are moving (or rather the system in general).

Let's discuss the forces. Yes, there is a force emanating from the fixed charge that is in the opposite direction of movement. But there is another force, at least as large, coming from you! Your force wins because you are able to push the charge forward. So both F and s are in the same direction. You have done positive work on the charge, and it has gained positive energy.

Switching to the case where the charges are oppositely charged, there is an attractive force between the charges. In this case, the charge is pulling you along with it. Think of a big dog on a leash that pulls you along with it. The work that you try to do pull back is futile. The sytem is doing work on you. In this case, your pulling force is in the opposite direction the charge ends up moving (i.e. your force loses this round). Thus the work that you do on the charge is negative. In other words, the system has done work on you, thus the system has lost potential energy.

In (2), the answer is 4.38 x 10^-17 J. However, I do not understand why it is a positive answer. Shouldn't the work be negative if the electron is being displaced against the electric field force?
Again, put yourself in there. If you are the one pushing the electron, and you give energy to the system, you have done positive work. (If work was easy, it wouldn't be called "work".)

Disclaimer: All of the above ignores ignores friction and such. It applies to conservative systems only.
no_audio
#3
Feb26-10, 10:22 PM
P: 4
I wish I could tell you how relieved I am to finally understand something that has been plaguing me for quite some time. I almost teared when I read the post. Thank you so, so much.


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