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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.

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  • $\begingroup$ The statement you cite is about a net opposing force, not individual forces. $\endgroup$ – ACuriousMind Oct 14 '14 at 10:51
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Oct 14 '14 at 11:12
  • $\begingroup$ How can one possibly know you have in mind non-conservative forces? $\endgroup$ – user59485 Nov 6 '14 at 9:58
  • $\begingroup$ @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. $\endgroup$ – user36790 Nov 6 '14 at 10:21
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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.

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    $\begingroup$ 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??? $\endgroup$ – user36790 Oct 21 '14 at 8:44
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    $\begingroup$ 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. $\endgroup$ – Floris Oct 22 '14 at 4:53
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    $\begingroup$ 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. $\endgroup$ – user36790 Oct 22 '14 at 5:46
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    $\begingroup$ 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 $\endgroup$ – bobie Oct 24 '14 at 12:40
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    $\begingroup$ @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 !! $\endgroup$ – user36790 Nov 5 '14 at 7:29
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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 $F < 0$.

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  • $\begingroup$ 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?? $\endgroup$ – user36790 Oct 14 '14 at 13:31
  • $\begingroup$ @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... $\endgroup$ – Floris Oct 14 '14 at 13:33
  • $\begingroup$ @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. $\endgroup$ – Johnny 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

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  • $\begingroup$ 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! $\endgroup$ – user36790 Nov 4 '14 at 7:51

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