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A body must do work against an opposing force to continue motion.

I have found this statement many times. But what is the reason behind it? Suppose $F_1$ is acting on a body to accelerate it (to increase its KE). Then another force $F_2$, less than the former, acts on the body in the opposite direction. So, according to the above statement, the body must have to lose energy.

But why will the body lose energy? Is it rather a rule or is there any logic? Now if $F_2$ becomes much greater than $F_1$, what will happen? Will the body still lose energy? Please give me logic so that I can understand this.

bobie
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  • The statement you cite is about a net opposing force, not individual forces. – ACuriousMind Oct 14 '14 at 10:51
  • If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Oct 14 '14 at 11:12
  • How can one possibly know you have in mind non-conservative forces? –  Nov 06 '14 at 09:58
  • @Alba: It doesn't bother the point if one thinks about conservative force. It is valid for both! Only that the in former, the lost energy is stored as PE in the system. –  Nov 06 '14 at 10:21

3 Answers3

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When two bodies interact, there is a force between them. Positive work is done on the object for which the dot product of force and velocity is positive. It follows that negative work is done on the object for which the dot product is negative.

By Newton's law (for each action there is an equal and opposite reaction), the force on one object is the reverse of the force on the other object - so necessarily if work done on one is positive, work done on the other is negative.

This is really just a statement of conservation of energy.

Floris
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  • I have some difficulty understanding the first line of your comment but I believe the answer is "yes, that's right". – Floris Oct 18 '14 at 09:33
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    Your answer is indirectly depicting the first law of thermodynamics,right?....when a body moves against a rough floor, kinetic friction $f_k$ opposes the body;the body will also exert same reacting force to the floor;but if the floor is not displaced,how can the body does positive work on the floor??? –  Oct 21 '14 at 08:44
  • Why do you say the floor is not displaced? The entire world is just a large sphere - angular momentum of that system has to be conserved. The motion is small because the mass is large... Of course part of the work is converted to heat. – Floris Oct 21 '14 at 09:27
  • Oh! You r right...part of the work done goes to rotate this earth's portion & part of it goes to increase the kinetic energy of the molecules,ie. thermal energy. –  Oct 21 '14 at 10:37
  • Sir I urge you to please answer...your words have worked greatly to clear my confusion...I request you earnestly to please answer this...it will be very grateful... –  Oct 22 '14 at 03:53
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    It is hard to answer your follow up comment - perhaps you should consider formulating it as a completely new question. It will get broader visibility and better responses. – Floris Oct 22 '14 at 04:53
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    thanks for the response,sir. I have been dying for your response. I have posted a question ; the answers I got haven't pointed at my problem & now gave a bounty to it. If you answer this or atleast write a comment, I will be very grateful,sir. Thanks for your valuable response. –  Oct 22 '14 at 05:46
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    Hi, Floris, @user36790 inquired after 'a body must...' and you replied: "When two bodies interact..." describing a totally different scenario, probably you added a lot to his confusion – bobie Oct 24 '14 at 12:40
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    @bobie maybe I did. But EVERY interaction involves two bodies - otherwise it isn't an interaction, by definition. – Floris Oct 26 '14 at 08:32
  • "Tomato falling from the sky" - in this question? I'm not seeing it. At any rate that scenario is still an interaction between tomato and earth. Two bodies. I am not getting the impression that I can be helpful here which is why I stopped responding. – Floris Oct 26 '14 at 10:40
  • He used a 'tomato' in a previous question. In the example here there is a body moving, say, a car, and a force, say ,air drag , opposing its motion. There are not 2 bodies, but 2 opposing forces on a single body. The formula is not 3rd law but Fdcosy. Another element of confusion might be: "By Newton's law ( I suppose you mean..3rd law)...This is really just a statement of conservation of energy." 3rd law is about conservation of momentum, in inelastic interaction ot 2 bodies energy may not be conserved. I hope I have been of help – bobie Oct 27 '14 at 10:25
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    As far as interaction is concerned, two bodies are always necessary. The agent that does positive work on the car is the engine. When air exerts opposing force to the car & does negative work, the car by virtue of 3rd law exerts force on the air and does positive work on it. So, Newton's 3rd law still applies! –  Nov 04 '14 at 08:03
  • @user36790 I think you've got it! – Floris Nov 04 '14 at 14:25
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    Sir,regarding what??? ...whatever it may be, i again thank you a lot for writing this life changing answer!! –  Nov 04 '14 at 14:33
  • I was responding to your earlier comment about interactions with two bodies, work done on air etc. It told me that your confusion had been cleared up which made me happy. – Floris Nov 04 '14 at 14:54
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    @Floris: In a word, your answer was the sunshine of enlightment amist those dark days! I don't want to rebuke anyother users but no one did ever point out the correct one. Thanks again and I hope you will help me again if I come to any problem! If there were chance of multiple votes, I would give you +1000 !! –  Nov 05 '14 at 07:29
  • @Floris: Horrendous, sir! Today, all on a sudden my quo & your ans got 2 downvotes. Why? No reason. It's really frustating to see an appropriate ans getting downvoted without the cause being written. Really abusive:( –  Mar 28 '15 at 16:22
  • @user36790 - downvoting is a valid means by which users on this site can disagree with an answer. The idea is that "good" answers will bubble to the top, although that isn't always the case. Lose no sleep over it. In the grand scheme of things eight points (four down votes on this answer!) is not a terrible loss. It is always preferable when people who disagree with an answer write a comment, but the site does not require it. People can downvote "just because", if they feel like it. Have a good day. – Floris Mar 28 '15 at 16:38
  • Sir, you are very modest:) . But that doesn't change the reality that it's very disgusting to get downvotes without knowing the cause. Was that a bad answer? Was there any flaw?? Not at all! I've no problem with downvoting, but a comment is worthy too. –  Mar 28 '15 at 17:54
  • I suspect it was a reaction to another answer - that sometimes happens. You need a certain amount of thick skin to survive on the Internet. I agree with your sentiments but recommend to "let it go". You will be happier for it. – Floris Mar 28 '15 at 17:55
  • @user36790, You did not get 4 downvotes as Floris says, neither 2, as you say. It was someone who had upvoted the question and then has reversed it to a downvote. It is not likely a reaction. This post was cited here when Floris was debating the issue in my bounty question (in which you deleted an answer). Floris was invited to refute the post, but declined. People come here to check the citation, and express their view. I downvote it when I deleted my post. I undeleted it now –  Mar 30 '15 at 09:01
  • @user36790, Probably one day, you too, like the one who recently , after a long time, realized his mistake, will realize that this post is wrong (and has misled you through half a dozen of follow-up questions), as everybody was trying to tell you. If you make even a minor edit, I'll be able to reverse my downvote to an upvote, and make it up for you: I was upset because you were offensive with the pittance of an upvote. But I see that your sunshine of enlightnment has already consoled you with an upvote elsewhere. :). I am upset now because Floris has done it again, spoilt my second bounty. –  Mar 30 '15 at 09:19
  • @GreenRay I didn't say his question got four downvotes - I said my answer did. I don't know what "second bounty" I spoilt. I am sorry that my presence and activity on this site seems to be a source of annoyance to you. – Floris Mar 30 '15 at 11:32
  • @Floris, actually I think you are a nice guy. But momentum is not your cup of tea and you keep posting wrong answers. You do not realize the impact you have on people. You see how devote this member is, that is exactly what has happened in the bounty question last week: they blindly believed you. You declared linus' post incomprehensible, posted a copy of ADG's post, and everybody voted your answer and ignored the good one. When someone has a high rep and many 'followers' should be more careful and considerate. I have probably been too harsh and I regret it, but I was furious last week. –  Mar 30 '15 at 13:25
  • I do not want to cause offense. While I don't believe my answer to the rod-on-ice question was in any way wrong, I will delete it to purge the bad taste out of everyone's mouth. – Floris Mar 30 '15 at 13:30
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If multiple forces act on a body at the same time, you should compute the net force before considering whether it is increasing or decreasing the kinetic energy of the object, and thus whether the work done by the force is positive or negative.

So, using your example, the net force is $F = F_1 + F_2$, and since $F_1 > 0$, $F_2 < 0$ and $|F_2|< F_1$ then $0 < F < F_1 $ and so, if the object is moving along the positive axis and $\Delta x > 0$, the work $W = F \cdot \Delta x$ done is positive.

Now, if the two forces don't act simultaneously then the work will first be positive, as both $\Delta x > 0$ and $F > 0$, then negative, since now $\Delta x > 0$ but $F < 0$.

Jessica Hansen
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  • I have asked why it is always a body have to lose energy when acted by an opposing force. What is the intuition behind it?? –  Oct 14 '14 at 13:31
  • @user36790 - conservation of energy. If you are pushing against an opposing force, the thing applying the force is having work done on it (since it feels the opposite force by Newton's law). Which means you must be losing it or energy is not conserved... – Floris Oct 14 '14 at 13:33
  • @user36790 I don't understand why you wouldn't find that intuitive. Remember that $F = ma$, so if the velocity is positive and the force is negative, the equivalent acceleration is negative. Thus a negative force will reduce the speed and thereby the energy. – Jessica Hansen Oct 14 '14 at 13:36
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Suppose $F_1$ is acting on a body to accelerate it (to increase its KE). Then another force $F_2$, less than the former, acts on the body in the opposite direction. So, according to the above statement, the body must have to lose energy.

But why will the body lose energy?

The logic you are looking for is not third law, it is called negative work

enter image description here

and the formula is

From the definition of the dot product, we have $$ W = F\Delta x\cos\theta $$ Where $F$ is the magnitude of $\mathbf F$, $\Delta x$ is the magnitude of $\Delta \mathbf x$, and $\theta$ is the angle between $\mathbf F$ and $\Delta\mathbf x$. Note, in particular that the magnitudes are positive by definition, so the $\cos\theta$ is negative if and only of $\theta$ is between $90^\circ$ and $180^\circ$. ... and the component opposite the motion contributes a negative amount to the work.

Therefore, if the angle is 180°, we have: $$ W = > F* d * -1 = (\cos 180°) $$ When we lift a box with an upward force F > g , g does negative work W -10m N on the box, and the net force is the difference F - W, according to vector addition

  • Yes, I know now it is negative work; I want to clear one thing: I have not made any mess with Newton's 3rd law . I will upvote you but can't accept it because it was Floris who enlightened me. Thanks! –  Nov 04 '14 at 07:51