# Potential energy and work

1. Sep 23, 2013

### ViolentCorpse

Hello everyone!

When we do work against a field on an object, why must the force we apply on it be equal and opposite to that of the field? If the object is initially at rest, then shouldn't we be applying a force greater and opposite to that of the field to first set it into motion? We could later make it equal after the object has been set into motion, but the first initial push should be greater.

Thank you! :)

2. Sep 23, 2013

### Andrew Mason

You are quite right. If the applied force was always equal and opposite, there would be no net force and, so, no acceleration. The applied force has to be just a tad greater initially in order to give the body some velocity and then it can be reduced to be equal in magnitude to the field force.

AM

3. Sep 23, 2013

### A.T.

That extra force needed for acceleration goes into kinetic energy, which is released when the object stops again. For "work against a field" we usually consider just the potential energy part.

4. Sep 23, 2013

### san203

If we want to do work against the field,the force must not be equal to that of field. We need to apply a force greater than what the field exerts on the object

True

Last edited: Sep 23, 2013
5. Sep 23, 2013

### Andrew Mason

That applies only to an object initially at rest. In general, positive work is done on a body by an applied force against the field force if the direction of the applied force is opposite to the direction of the field force and if it is applied through some positive displacement. It doesn't necessarily have to be greater in magnitude than the field force.

AM

6. Sep 23, 2013

### ViolentCorpse

Shouldn't the entire journey of that object against the field comprise of part potential and part kinetic energy?

Thank you so much each and everyone of you!

Last edited: Sep 23, 2013
7. Sep 23, 2013

### Andrew Mason

If you meant that the work done by an external force, F, in moving the body from point a to point b: ($\int_a^b \vec{F}\cdot d\vec{s}$) must be equal to the sum of the change in kinetic energy + change in potential energy of the body from a to b, then the answer is: yes.

AM

8. Sep 24, 2013

### ViolentCorpse

Thank you for clearing that, though I know that much. I'm just trying to see how accurate it is to express work done against the field as F.ds, if that F was a little more than F for a very short period of time to get the object moving in the first place..

9. Sep 24, 2013

### A.T.

What is F here?

If F is the negated field-force then integral F.ds is the work done against the field.

If F is the net force, then integral F.ds is the net work done.

10. Sep 24, 2013

### Andrew Mason

If Fa is the applied force and Ff is the field force, the work done by the applied force on the object in moving from a to b is: $\int_a^b \vec{F_a}\cdot d\vec{s}$, which consists of the change in potential energy + change in kinetic energy. The work done against the field is $\int_a^b \vec{F_f}\cdot d\vec{s}$, which is just the change in potential energy.

AM

Last edited: Sep 24, 2013
11. Sep 24, 2013

### ViolentCorpse

Oh, I think I get it now. I'm clearly not a very bright student.

Thank you, Andrew Mason and A.T for your continued help. I very much appreciate it.

12. Sep 24, 2013

### Andrew Mason

On the contrary. You asked a very good question. This is something that many physics texts gloss over. You learn by asking questions. The only kind of dumb question is one that should have been asked - but wasn't.

AM

13. Sep 26, 2013

### ViolentCorpse

That's really very nice and encouraging of you, Mr. Andrew Mason. Thank you!