# Is work done against a field when the forces match?

I was trying to get my head around the energies of particles and anti particles in the same force field when I got a bit off topic.

As I understand it, work = force * displacement, where the displacement is due to the force acting on it. However imagine I have a mass which starts at some velocity Vi in the direction against gravity.

Now imagine I continually supply a force against gravity equal to the weight of my object. In this scenario my mass doesn't accelerate and it's energy is just it's potential due to its height.

But is work done in this scenario? At each point Fgravity = Fsupply, and so displacement is only due to my initial velocity Vi. So does my definition of work need some work, or am I missing something?

• Does this post help? physics.stackexchange.com/q/295245/104696 – Farcher Dec 1 '16 at 0:33
• That post would imply work isn't done on the mass in this scenario? Yet the potential energy of the mass increased, and energy has been put into the mass for it to be able to move against the gravitation field – user45815 Dec 1 '16 at 0:50

Because of your comment I want to explain further but cannot do it as a comment as my explanation is too long.

That post would imply work isn't done on the mass in this scenario? Yet the potential energy of the mass increased, and energy has been put into the mass for it to be able to move against the gravitation field.

Let the object under consideration be a point of mass $m$ and the other possible participants are the Earth and you.

The gravitational field strength is $g$ and the object is lifted (moved away from the Earth) by you a vertical height $h$.
At the start the object is not moving and at the finish the object is not moving.

Work done is the dot product of force and displacement.

If the displacement is in the same direction as the force then the work done is positive and if the displacement is in the opposite direction to that of the force the work done is negative.

System = object alone

There are two forces acting on the object.
The force on the object due to the gravitational attraction of the Earth (weight) of the object downwards $F_{\rm grav} = mg$ and the the force that you are exerting on the object $F_{\rm you} = mg$.

Work done by you on the object is $+mgh$
Work done by the gravitational field on the object is $-mgh$
Total work done on the object is $+mgh +(-mgh) = 0$

This could also have been worked out by saying that the net force on the object is $+mg+(-mg)=0$ and so the wrk done is zero.

Having considered the work done on the object what can we say about the box.
It has moved a distance $h$ away from the Earth and having started with no kinetic energy it finished up with no kinetic energy.

At that is about it because $F_{rm grav}$ and $F_{rm you}$ are both forces which are external to the box.
Gravitational potential energy is a property of the Earth and the object and we have only considered the object alone.

System = object and Earth

In this case you are applying the only external forces.
It is forces plural because you need to separate the object and the Earth by a distance $h$.
Think of a spring being stretched.
To stretch the spring two equal in magnitude but opposite in direction forces must be applied on the spring.

Assume that the mass of the Earth is very much bigger than that of the object then the work that you do on the Earth-object system is still $mgh$.
In this Earth-object system the gravitational forces on the Earth and the box are internal forces so are not included in the work done by external forces sum.
The object started with no kinetic energy and finished with no kinetic energy so where has the work that you did go?
The answer is that is has been used to increase the gravitational potential energy of the $Earth-object$ system by $mgh$.

What I have tried to do is to show you that the choice of system is vital and in this case the object alone cannot have gravitational potential energy it is the Earth-object system which has the gravitational potential energy.

The reason that this point is often missed or ignored is that in the Earth-object system the Earth has a “passive” role other than producing a gravitational field because its mass $M$ is so much greater than that of the object $m$.
When you apply those external forces on the Earth-box system to increase the separation of the Earth and the object by $h$, relative to the centre of mass of the system the Earth moves a distance $\dfrac{mh}{M+m}\approx 0$ and the object moves $\dfrac{Mh}{M+m}\approx h$.
If you drop the object then the momentum gained by the Earth (equal and opposite to the momentum gained by the object) is ignored because it is the speed and kinetic energy of the object which is of interest and those quantities are so much greater for the object than for the Earth.
So you do not write potential energy lost by the Earth-object system is equal to the kinetic energy gained by the Earth plus the kinetic energy gained by the object because the kinetic energy gained by the Earth is so much smaller than that gained by the object.