Work done by conservative force

1. Nov 14, 2015

gracy

I know work done by conservative forces= - $ΔU$=$ΔK.E$
But I have a question.Does it mean work done by all the conservative forces present (in a particular physics problem)= - $ΔU$=$ΔK.E$
or just work done by a conservative force= - $ΔU$=$ΔK.E$
I mean let's say a problem in physics involves both electrostatic and gravitational force We know both are conservative forces.Should I equate work done by gravitational force +work done done by electrostatic force=-$ΔU$=$ΔK.E$
or should I treat each force separately.If yes,then I don't now how am I going to treat each force separately.

2. Nov 14, 2015

Matterwave

It should be the total.

3. Nov 14, 2015

vela

Staff Emeritus
You should consult your textbook and see how this equation is derived. That should provide answers to your questions.

4. Nov 14, 2015

gracy

You mean this one is correct
work done by gravitational force +work done done by electrostatic force=-$ΔU$=$ΔK.E$

Last edited: Nov 14, 2015
5. Nov 14, 2015

gracy

My text book does not involve the derivation, it just includes application .

6. Nov 14, 2015

haruspex

I confirm it is the total. Think of it this way: the object experiences a net force $F_{net}$. We know that determines its motion: $F_{net}=ma$. The acceleration, over a period, determines its change in KE.
By the way, we are here discussing the conservative forces acting on the body which gains this KE. We are not interested in forces acting on other bodies in the same system.

7. Nov 14, 2015

gracy

You mean this one is correct
work done by gravitational force +work done done by electrostatic force=-$ΔU$=$ΔK.E$
And similarly in case of non conservative force
Work done by non conservative force=$ΔK.E$+$ΔP.E$
Does it mean Work done by all the non conservative forces present in a (particular physics problem)=$ΔK.E$+$ΔP.E$

8. Nov 14, 2015

vela

Staff Emeritus
I doubt that, or you have an incredibly bad textbook.

The work-energy theorem comes from integrating Newton's second law:
$$\int_{\vec{r}_i}^{\vec{r}_f} \left(\sum_i \vec{F}_i\right)\cdot d\vec{r} = \int_{\vec{r}_i}^{\vec{r}_f} m\vec{a}\cdot d\vec{r}.$$ The righthand side is equal to the change in kinetic energy. On the lefthand side, we can pull the summation out of the integral to get
$$\sum_i \int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_i \cdot d\vec{r} = \frac 12 mv_f^2 - \frac 12 mv_i^2 .$$ Each integral in the summation is just the work done by force $i$, so we have
$$W_1 + W_2 + W_3 + \cdots + W_n = \frac 12 mv_f^2 - \frac 12 mv_i^2$$ where $W_i = \int \vec{F}_i\cdot d\vec{r}$. If a force is conservative, we bring its contribution to the work over to the other side of the equation and rename it as the change in potential energy. That is, $\Delta U_i = -W_i$. The stuff left on the lefthand side is the work done by non-conservative forces. So now we have
$$W_{nc} = \Delta KE + \Delta U_1 + \Delta U_2 + \cdots + \Delta U_k.$$ If there are no non-conservative forces, the lefthand side is 0, and you end up with
$$0 = \Delta KE + \Delta U_1 + \Delta U_2 + \cdots + \Delta U_k.$$

9. Nov 14, 2015

haruspex

I think you misunderstand the significance of "conservative" here. The work done on an object by any force is ΔKE. What's special about conservative forces is that the work done is independent of the path taken. If the object returns to its original position, the total work done by the force is zero.
No. PE is not a property of the individual body. It is a property of a system. For gravitational PE 'of' an object in Earth's field, the PE is possessed by the object-Earth system. A force may do work which raises that PE, but is not considered work done on the body.
A force, conservative or not, may do work on the body (ΔKE) and work on the system (ΔPE).

10. Nov 16, 2015

gracy

Here also I categorized conservative forces ,If it is correct
why not this
work done by Sum of all the non conservative forces=ΔK.E+ΔP.E

11. Nov 16, 2015

haruspex

Why do you keep referring to conservative and non conservative forces in this thread? I have mentioned several times that these work equivalence equations have nothing to do with whether the forces are conservative. I have provided you with the condition that determines whether a force is conservative or not.

12. Nov 19, 2015

Staff: Mentor

I agree with Haruspex. The source of your confusion lies in the practice of referring to forces as conservative and non-conservative. Whoever taught it to you this way has done you a disservice. How about revealing the name of the book you are using so others can be warned against using it.

Chet

13. Nov 19, 2015

gracy

14. Nov 19, 2015

Staff: Mentor

For whatever it's worth, I think you are obsessing over this terminology too much. Your time is much too valuable for that. If you want to specify a specific problem, we can talk about how to solve it.

Chet

15. Nov 19, 2015

gracy

But It s much convenient for me to use those terminologies (conservative /non conservative)I loved the answer Mister T gave me in https://www.physicsforums.com/threads/definition-of-potential-energy.834808/page-4 post #61/I was having only one doubt which I posted here in this thread but now all are telling me to stop using those terminologies .I desperately want to stick to those terms and formulas (conservative/non conservative)Can not I?Is there no way?

16. Nov 19, 2015

Staff: Mentor

You should do whatever works best for you. It's a personal thing.

Chet

17. Nov 19, 2015

gracy

You mean it is not wrong to use conservative/non conservative forces for work calculations?

18. Nov 19, 2015

Staff: Mentor

Like I said, whatever works best for you.

19. Nov 19, 2015

gracy

Then I will choose conservative/non conservative force approach for work done calculation,but as I said to apply those formulas I need to clear a doubt