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Let us consider 2 objects A and B.

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The initial PE of the objects be U(A)=10J and U(B)=0J. Let F be the force exerted by A on B. Suppose W be the work done by force F. Due to this the PE of both the objects will change to U'(A)=6J and U'(B)=4J.

Work Done by A=Energy Released by A

W=ΔU(A)

W=U(A)-U'(A)=10-6 W=∆U(A)=4J

Work Done by A=Energy Gained By B

W=-ΔU(B)

W=-[U'(B)-U(B)]=-[4-0] W=-∆U(B)=-4J

ΔU(A)=-ΔU(B) ΔU(A)+ΔU(B)=0 implying that

Change In PE of A + Change in PE of B=0

If i am not wrong here till now then there is a doubt in my mind(If i am wrong somewhere above then tell me my mistake) The doubt is

According to conventions, Energy given out is -ve and Energy gained is +ve in Physics. But in my answer,

W=ΔU(A)=Energy Released by A=4J which should be -4J according to conventions

W=-ΔU(B)=Energy Gained by B=-4J which should be +4J according to conventions

So my doubt is why conventions related to work here isn't going right?

Perspicacious
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    I think you are having more trouble with mathematics here than with physics. To denote that a quantity is positive we write $x>0$, rather than $+x$ and for negative quantities we write $x<0$ rather than $-x$. The easiest way to get all of this right is to use a summing notation: $E_A + E_B =0$ or, if you feel uncomfortable setting the absolute value to zero, you can use $E_A + E_B=E_{total}$, but the latter really doesn't mean much more, as the total energy of a system is observer dependent and for your example only the energy transfer accounting matters. – CuriousOne Jul 03 '16 at 19:02

1 Answers1

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When you write "energy given out" or "energy gained" you are expressing a choice of sign for the energy transfer that you are specifying, and each of them is different.

Notice that in your scenario A loses energy, so it's "energy given out" is positive, but it's "energy gained" would be negative.

Employing a sign convention means always stating changes of energy as if one of those two phrase was implied. That is, when you use a sign convention you don't use the phrase "energy gained" or "energy given out" because they are built in. The convention you state (and it is not universal, I'm afraid, different books make different choices) is the one of stating changes of in terms of "energy gained".

Then we come to your calculations.

You start by computing the initial energy for A minus it's final energy. Which is the energy given out (and so has the opposite sign convention of the one you said you were using).

Then you compute the energy change of B by asserting (correctly) that it is the opposite of that of A. But because you did A in the wrong convention, your result for B is also in the wrong convention.

BTW, not only do different authors make different choices for the sign convention, but some make different choices depending on the kind of energy transfer. It is fairly common for engineering treatments to set heat positive when the system gains energy and work positive when the system loses energy; making the first law of thermodynamics $\Delta U = Q - W$. Presumably because the 18th and 19th century engineers were interested in shoveling coal into their engines and getting work done as a result.

dmckee --- ex-moderator kitten
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  • I didn't get ur point that "A loses energy, so it's "energy given out" is positive, but it's "energy gained" would be negative".Can you explain it? Also what your post is saying is that I took a wrong convention for A. Should I take W=U'(A)-U(A)=-4J. If this what u are saying then W=∆U(B) which is not true for any conservative force. Can u please explain all this? Thanx for the time and effort u gave in this thread – Perspicacious Jul 06 '16 at 07:23