# Conservation of Energy

1. Oct 14, 2015

### Devilwhy

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
4a at x=-2 F=0N
Potential energy=integration of F(-2)=Integration of 0=0 so it will reach minimum at x=-2?
b. integration of F(-2)-integration of F(-2-h)=m(v-0)^2/2
am i right?

5 Centripetal force=mv^2/r=mgsinθ-Fn(normal force)
will the block fall leave surface when Fn=0?

so i can represent the velocity as v=(mgrsinθ)^1/2=((m)(9.8)(15)sinθ)^1/2?

2. Oct 15, 2015

### Simon Bridge

Force F is the gradient of the potential energy U.
A function has an extremum when its gradient is zero.
Therefore U has a minumum at some position where F=0

Integrate F(x) to get the PE function U(x), then put in x=-2 to find U(-2) $\neq$ 0

Also... what is U(x0 - h)?

How do changes in kinetic energy relate to changes in potential energy?

3. Oct 19, 2015

### CrazyNinja

Your attempt at the first question [4(a)] looks correct. As F=0 at x= (-2), it is the co-ordinate where U is minimum.

There are complications in 4(b) though. the work done by the new force cannot be determined from given data. It renders the Law of Conservation of Energy useless. How then do we proceed?

For 5, use the Law of Conservation of Energy. That should give you the required answer.

4. Oct 20, 2015

### Simon Bridge

I can do it from the data supplied, using conservation of energy.
Remember the relationship between kinetic and potential energy.

5. Oct 20, 2015

### CrazyNinja

The new force is not given to be conservative. Hence the work it does requires the path too, of which nothing is mentioned. Nor is the equation governing the magnitude.

6. Oct 20, 2015

### Staff: Mentor

The new force is not acting when the object is released. The new force only served to move the block to its new location and plays no role in the work done by the conservative force when it is released from rest from that location.

7. Oct 21, 2015

### CrazyNinja

What you have said is indeed correct. But in order to determine the initial "total" energy of the object, we require the work done by the new force. I am talking about the process in which the new force moves the particle form x0 to x0-h. KE is zero in both the cases. Initial PE is U(x0). Final PE is U(x0-h). In addition to this, there is additional work done by the new force. Thus, U(x0-h) = U(x0) + W

If this work done is known, then we know the total energy of the particle at x= x0-h. Only then can we proceed with its release and use this to calculate max KE which, interestingly, will be equal to the work done by the new force.

8. Oct 21, 2015

### Staff: Mentor

No. The history of the object is totally irrelevant. Additional work would show up as KE, but the object is at rest at the new location. Whatever work was done to move the object left no evidence of the process. In a conservative field PE is determined by location alone.

No, no, no. The only incarnations of energy in this system are KE expressed as motion and PE as a result of location in the field. Objects don't otherwise "remember" their history. Once an object is brought to rest there's no inherent evidence that the object experienced any particular process or path.

9. Oct 21, 2015

### CrazyNinja

And what you have said is again true, though in this case there are a few additional points. PE is determined by location alone in a conservative field, but external work is not. The Law of Conservation of Energy states- " E(final)= E(initial) + W(external)", where I have included PE in E() and W(external) implies work done by non-conservative forces, which I have excluded from my system. In this context, the object remains at rest because the external work done (by the new force) is equal in magnitude to the change in PE and opposite in sign, which is what U(x0-h)= U(x0) + W means.

I disagree with you. The work done by the new force will also manifest as a form of energy and will play a role in the "energy" equation.

10. Oct 21, 2015

### CrazyNinja

Okay. I just realised something.

I mentioned the answer in my own post and was arguing about it.

This basically implies that change in PE is the max KE. Im sorry for the inconvenience caused.

This brings a question to my head: Was I right all this while, or were you guys right all the time, or were both of us right and arguing for the same thing (which I dunno why happens a lot with physics -_- )??

11. Oct 21, 2015

### Staff: Mentor

Does this energy show up as KE? Does it show up as PE? Something else? Will two objects brought to rest at the same location by different means or routes have different energies? What test could be performed on such objects to tell them apart?

12. Oct 21, 2015

### CrazyNinja

I guess your post answers this the best. Which means I was wrong.
I still do not understand where I went wrong though. The equations I wrote look consistent. In addition to that, they tell me that the new force is one with similar characteristics to the conservative force whose field we are working in.

13. Oct 21, 2015

### Staff: Mentor

No worries. What's important is that you're thinking about the physics and working your way to understanding. One aha! is worth a week of memorization
What you wrote said that the total work done to move an object from one location to another in a conservative field is equal to the change in PE between those locations. That's fine! In fact it's the definition of PE for a conservative field. Where things went awry was in deducing or perhaps implying that the method or route taken would leave an "imprint" on the result. It is a common misconception that crops up when conservative forces and fields are first introduced.

14. Oct 21, 2015

### Simon Bridge

The trick is to think about what sort of energy other than PE or KE could be involved ... how would that energy, under the condition of free motion in the PE field, affect the resulting kinetic energy? The extra W in your equation - by what mechanism would it turn into KE?

$U(x_0-h) + W = K_{max} + Q$ ... by what mechanism would $W \neq Q$ that is consistent with the description and context?
(The object is held at rest and then released.)

In short - it is not so much that you were wrong exactly, but that you were over-thinking the problem.
Implicit in any physics problem is that the problem should be solvable, in a sensible way, by the average student having completed and understood the coursework ... physics problems never have all the information included explicitly or they would be very long-winded. Part of doing the problem is making a value judgement about what information is sensible to be included.. part of learning physics is learning to make that judgement.

In this case, unless the student has other information, we can consider that everything we need to know about the effect of the additional non-conservative force has been provided: that it's result is to place the object, initially stationary, other than at the PE=0 point.
Other possible effects depend on information not supplied and not relevant to the lesson - so: it's a red herring.