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How to understand the work-energy theorem?

I took a short lecture on physics for engineering last week. The lecturer emphasized that the work done on an object will cause the kinetic energy change as

$$W = \Delta \text{KE}.$$

I know this concept might be so common to you but to me, as a beginner, it is pretty hard to understand the reason. My understanding is that 'work' is the energy an external object 'injects into' the object or is the energy an external object 'takes away' from the object. I think the work done by the object should equal to the total energy changed on that object, which could be in any form (heat, potential or kinetic energy.) Why does the theorem only explicitly refer to kinetic energy? Will this theorem work in some cases or in all cases?

dmckee --- ex-moderator kitten
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user1285419
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2 Answers2

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The total work can be split up into two parts:

$$W_{net} = W_{conservative}+W_{non-conservative}.$$

With the conservative part you can associate a potential energy:

$$W_{conservative}=-\Delta PE$$

(this is in fact the definition of a conservative force) so that the Work-Energy theorem becomes

$$W_{non-conservative}=\Delta KE + \Delta PE = \Delta E.$$

This is another way of writing the Work-Energy theorem and in my mind it's a little bit clearer. Restated, the work done by non-conservative forces is equal to the overall change in energy of the system.

For example, work done by friction is negative, so it dissipates energy away from a system.

On the other hand, gravity is a conservative force. Imagine the motion of a falling ball. Unless something doing work on the ball to slow it down (for example, air) the ball will speed up as it falls. In this case, the equation

$$W_{gravity} = -\Delta PE = \Delta KE$$

is equivalent to that statement. (As the potential energy becomes more negative, kinetic energy becomes more positive.)

Pricklebush Tickletush
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  • thanks Alec. I am still a bit confusing on that. From your statement, should I say the conservative force will only change the potential energy but not the kinetic energy? I know that gravity is a conservative force, so if I freely drop an object from high place (ignore the friction), the gravity will do the work on the object so to lower the potential energy but its speed will change too (so does the kinetic energy). So how to understand this? – user1285419 Mar 26 '13 at 20:42
  • I've been editing this to make it more clear. I have added an equation to the end illustrating the connection between gravitation work, potential energy and kinetic energy. – Pricklebush Tickletush Mar 26 '13 at 20:46
  • by the way, in some books, some cases, I find that they will put the internal energy change in the work-energy theorem too. So in what situation I should consider the internal energy and what's the reason to cause the change of internal energy? If some case it said the internal energy of the system changed, can I said there must be some non-conservative force applied on the system? thanks – user1285419 Mar 26 '13 at 20:46
  • In the equation $W_{non-conservative}=\Delta KE + \Delta PE = \Delta E$ you can divide $\Delta E$ into as many parts as you like. It really depends on the context. If you're talking about a chemical reaction, for example, you may want to have an internal energy term, but not if you're talking about pushing blocks on inclined planes. – Pricklebush Tickletush Mar 26 '13 at 20:50
  • Thanks Alec, I think your explanation sort of giving me some points to start. Thanks again. – user1285419 Mar 26 '13 at 20:52
  • I know this questions was closed but I am a bit confusing because from several books I read yesterday, they always show the work-energy theorem as $W_\text{net}=\Delta KE$, so it gives me a feeling that no matter what the forces are (conservative or nonconservative), it is always true that the net work done by the net force equals to the change of kinetic energy. Is that really true? What confusing me is in the textbooks I am reading, none of them talk about something like $W_{non-conservative} = \Delta KE + \Delta PE$ – user1285419 Mar 28 '13 at 02:39
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    @user1285419 Ok sure. So imagine you throw a ball up in the air. You did work on it, so you increase its kinetic energy. But something happens as it rises -- it slows down! That's because gravity is also doing work on the ball -- negative work, in fact, so that the ball's kinetic energy goes down. But the difference between a force like gravity and one like friction is that you can easily get the energy back from gravity. That's why we call it a conservative force. – Pricklebush Tickletush Mar 28 '13 at 17:08
  • @user1285419 Now we could stay at the first statement of the work energy theorem, that is $W_{net}=\Delta KE$, or we could redefine the work done by gravity on the ball as a new kind of energy because we can always get that energy back, if we want to, by just letting the ball fall. It's this recoverable energy that we define as potential energy, $W_{conservative} = - \Delta PE$. (The minus sign comes from the fact that when you slow something down (do negative work) you increase this other quantity, the potential energy.) These statements are equivalent. – Pricklebush Tickletush Mar 28 '13 at 17:10
  • What will happen suppose, when initially there is a potential energy on a body, and as it is being converted to Kinetic energy, a frictional force opposes to it? How do I apply work energy principle in that case? – Aneek Sep 19 '15 at 06:40
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    This answer has problems. It does not address the issue of system and environment and does not distinguish between internal and external forces. This accounts for the confusion expressed in many comments. – garyp Jan 24 '21 at 14:58
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I think the problem with the Work-Energy theorem is that so few books properly cover the concept of defining a system. If we take the example of Earth and a ball near the surface of Earth as an example, we can ask "What is the total energy of the system?". Obviously, this is the sum of the kinetic energy of Earth, the kinetic energy of the ball, and the gravitational potential energy of the earth and ball. $$ E_\mathrm{tot} = K_{\mathrm{earth}} + K_\mathrm{ball} + U_\mathrm{earth-ball}$$

When we apply the Work-Energy theorem to the ball, the earth is no longer part of our system, and therefore the gravitational potential energy is also not part of the definition of the total energy of the system. $$E_\mathrm{tot} = K_\mathrm{ball} $$ Therefore, any change in the gravitational potential energy between Earth and the ball must be considered work being done on the system (which for the Work-Energy theorem is just the ball). Since the only energy term in the total energy in this case is the kinetic energy of the ball, the work done by a change in the gravitational potential energy between Earth and the ball must change the ball's kinetic energy.

scmartin
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