# B Gravitational potential energy conflict

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1. Aug 25, 2016

### donaldparida

I know that when an object A a does positive work on another object B, object A loses energy and object B gains energy(there is transfer of energy from object A to object B) and when object A does negative work on object B, it gains energy and object B loses energy(there is transfer of energy from object B to object A).
Now, when we lift an object(standing on the Earth), we do positive work on the object and so the energy of the object should increase (and our energy should decrease) while the Earth by its gravitational force does negative work on the object and so the energy of the object should decrease(and the energy of the Earth should increase).
Now my question is, is the energy of the object increasing or decreasing since it cannot be both at the same time.

2. Aug 25, 2016

### Staff: Mentor

It can be both at the same time. Think of a rope where one end of the rope is attached to a horse and the other end is attached to a cart. The horse is doing positive work on the rope, transferring energy into the rope, and the cart is doing negative work on the rope, transferring energy out of the rope at the same rate.

Similarly, it is perfectly acceptable for you to be transferring energy into a rock at the same rate as gravity is transferring energy out of the rock.

However, that is usually not the conceptual error that is being made. Usually the error is that the system under consideration is being changed. You can arbitrarily define the system, but then you have to be consistent.

One possibility is to define the system as the object+earth. In this case as you lift the object you transfer energy into the object+earth system, which is stored in the PE of the system. The gravity is an internal force, so it doesn't transfer any energy.

The other possibility is to define the system as the object only. In that case as you lift the object you transfer energy into the system. The gravity is an external force and the energy is transfered out of the system and into the gravitational field (fields store energy).

The confusion usually comes by considering the object+earth to be the system in one step and then accidentally switching to considering the object only system in the middle.

3. Aug 27, 2016

### donaldparida

If the object is now released (system: object only) then the gravitational field will do positive work on the object and the energy of the object will increase( and the energy of the field will decrease) and the object will fall down. Will the object have more energy now compared to the starting?

4. Aug 27, 2016

### Staff: Mentor

Sure. It's speeding up.

5. Aug 27, 2016

### donaldparida

What will happen to the energy of the object when it hits the ground?Will it be transformed to heat energy or remain with the object?

6. Aug 27, 2016

### Staff: Mentor

Assuming a typical object/ground collision--like dropping a book, for example--the kinetic energy of the book will end up as random thermal energy (for the most part).

7. Aug 28, 2016

### G. Cavazzini

Donaldparida wrote:
[...] when we lift an object(standing on the Earth),
OK
we do positive work on the object
These words are improper: in this case we do work against gravitation, not on the object.
and so the energy of the object should increase
No, if we exclude the transient we lift the object.
(and our energy should decrease)
Yes, we spend energy to lift the object, against gravitation.
while the Earth by its gravitational force does negative work on the object
No, not on the object. On us (on our hands and arms), who are lifting the body.
and so the energy of the object should decrease
The object has no (kinetic) energy, except in the transient we lift it.
The energy of the object does not decrease.

(and the energy of the Earth should increase).
No.
Now my question is, is the energy of the object increasing or decreasing since it cannot be both at the same time.
The body has kinetic energy only in the time-span you spend to lift it.

8. Aug 29, 2016

### donaldparida

9. Aug 29, 2016

### jbriggs444

You should learn to use the quote functionality built into these forums. When replying to a post, you can either hit "reply" and have the post to which you are responding surrounded in a quote tag and your cursor positioned ready for you to key in a response. Or you can highlight individual sections to which you want to reply, right-click and select "quote". Then, in your reply, click "Insert quotes". Now back to the matter at hand...

When we lift an object, we most certainly do work on the object. Whether a separate gravitational force extracts work from the object has nothing to do with it.

Our contribution to the objects's energy is positive. There are other contributions. If we count gravitation as a potential, the object's potential energy increases and its kinetic energy remains constant. The object gains energy. If we count gravitation as an external force, it subtracts from the object's energy. The object does not gain kinetic energy.

There may be other contributions to our own energy. For instance, if we are standing on an escalator holding a shopping bag, we can do positive work on the shopping bag without losing energy in the process. The escalator steps under our feet are doing positive work on us even while the shopping bag is doing negative work. In such a case, we are gaining potential energy in the process.

Last edited: Aug 30, 2016
10. Aug 30, 2016

### vanhees71

The energy-conservation law in the most simple case of a conservative force is immediately derived from Newton's equation of moation. A force is called conservative, if it is described by a scalar time-independent potential,
$$\vec{F}(\vec{r})=-\vec{\nabla} \phi(\vec{r}).$$
$$m \ddot{\vec{r}}=-\vec{\nabla} \phi(\vec{x}) \qquad (*).$$
Now multiply this equation with $\dot{\vec{r}}$ (scalar product). The left-hand side becomes
$$m \dot{\vec{r}} \ddot{\vec{r}}=\frac{\mathrm{d}}{\mathrm{d} t} \left ( \frac{m}{2} \dot{\vec{r}}^2 \right),$$
and the right-hand side
$$-\dot{\vec{r}} \cdot \vec{\nabla} \phi(\vec{r})=-\frac{\mathrm{d}}{\mathrm{d} t} \phi(\vec{r}).$$
Subtracting both equations leads, together with (*) to
$$\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{r}}^2+\phi(\vec{r}) \right )=0.$$
This means that the quantity in the bracket, called the energy of the particle, is constant, i.e.,
$$E=\frac{m}{2} \dot{\vec{r}}^2 + \phi(\vec{r})=\text{const}.$$
As an example take gravity of the earth on an object. Close to earth you can write the force as
$$\vec{F}=m \vec{g},$$
Obviously this is a conservative force with
$$\phi=-m \vec{r} \cdot \vec{g},$$
$$E=\frac{m}{2} \dot{\vec{r}}^2 -m \vec{r} \cdot \vec{g}=\text{const}.$$

11. Aug 30, 2016

Staff Emeritus

12. Aug 31, 2016

### donaldparida

What do you mean by B thread?

13. Aug 31, 2016

Staff Emeritus
The level of the thread - basic.

14. Aug 31, 2016

### vanhees71

Well, B thread means you have to explain something with so many words and so little math that it becomes incomprehensible even to the expert...

15. Sep 5, 2016

### hmmm27

Potential Energy, per se, is a relationship, not a solo act.

Within the context of the object itself, no internal energy is gained by raising it up a few feet : the molecules didn't wind themselves up tighter, the amount of heat being emitted (or absorbed) didn't change, etc.

Within the context of the object and the Earth, some Potential Energy is gained in the relationship (though we just refer to the object as owning it to make sentences shorter).

As an aside you could increase the PE by digging a hole in the ground under the object so it has farther to travel. Or, for fun, you could stand on your head, balance the object on your feet and lift the Earth, it works out the same, though the latter involves much longer numbers to get the same answer.

Help any ? or too basic.

16. Sep 8, 2016

### MicroCosmos

Sorry but i dont think so. When you release an object it transfers its potential to cinematic energy. it doesnt win anything.

17. Sep 8, 2016

### MicroCosmos

And sorry again but it cannot be both at the same time. You can get energy and give at the same time. The addition determines wheter you are losing, gaining, or remaining with the same energy.

18. Sep 8, 2016

### jbriggs444

19. Sep 8, 2016

### MicroCosmos

lol, i scrolled the whole page looking for what i was missing until i got it

20. Sep 8, 2016

### jbriggs444

I should apologize for poking fun at an unintentional mistake made by a non-native English speaker. The term is "kinetic" energy, not "cinematic" energy. Kinetic energy is the energy associated with motion. "Cinematic" has to do with motion pictures.