# What happens to energy when work is done?

Okay say I have an object with kinetic energy and it collides with a stationary object.

The moving object does work. Work means energy spent thru a force and distance right?

So my question is once the energy gets used, does the moving object stop becuase it has no energy. Or is it stopped because of Newton's third law. I know there is a reaction force but I feel one the energy is gone why would there be a reaction force.

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ZapperZ
Staff Emeritus
Okay say I have an object with kinetic energy and it collides with a stationary object.

The moving object does work. Work means energy spent thru a force and distance right?

So my question is once the energy gets used, does the moving object stop becuase it has no energy. Or is it stopped because of Newton's third law. I know there is a reaction force but I feel one the energy is gone why would there be a reaction force.
This is rather confusing to understand.

I'll ask you this. If the original moving object is being opposed by, say, your hand, i.e. your hand pushes it in the opposite direction that it is moving so that it is slowing down until it comes to a halt, do you still have a problem in understanding your original question?

Zz.

Yes I understand your example. However, now I apply work to that. Where did the energy go? Didn't the hand apply a force in an opposing direction and consume energy? What about the energy of moving object?

My question is simple. Work is measure of energy expended. If a moving object expends it's energy on a stationary object, wouldn't that stop it not the reaction force

I get work and such for most part. I'm confused how it works in an elastic collision where one object is moving with KE, another is not. After collision, they have switched energies

sophiecentaur
Gold Member
Where did the energy go?
My question is simple.
In fact it is not straightforward - except in as far as Energy is Conserved ( in non relativistic situations). In any physical interaction, some energy will end up as thermal (heat), some as Kinetic Energy and some as Potential Energy.
All situations are different but the above principle applies. Think of a situation and you can fill in the blanks yourself.

kuruman
Homework Helper
Gold Member
I get work and such for most part. I'm confused how it works in an elastic collision where one object is moving with KE, another is not. After collision, they have switched energies
This is how it works in the case of a moving object colliding with one at rest:
While the two objects are in contact
1. They are displaced by the same amount since they move together as one (that's what contact means).
2. The "projectile" object exerts a force in the same direction as the common displacement. Therefore, the projectile does positive work on the target.
3. The target object exerts an equal force but in the opposite direction (reaction counterpart) to the common displacement. Therefore, the target does negative work on the projectile.
4. The two works have the same magnitude because the forces have equal magnitudes at all times according to Newton's 3rd law.
5. The kinetic energy of the target increases because positive work is done on it while the kinetic energy of the projectile decreases because negative work is done on it.

Put it all together and you see that there is a net transfer of kinetic energy from the projectile to the target. If all the energy is transferred from the projectile to the target then we say we have an elastic collision. If friction, sound, etc. take their cut, we say we have an inelastic collision. The colliding objects will not always switch energies. For that to happen two conditions must be fulfilled: (a) the collision must be elastic and (b) the objects must have the same mass.

ZapperZ
Staff Emeritus
Yes I understand your example. However, now I apply work to that. Where did the energy go? Didn't the hand apply a force in an opposing direction and consume energy? What about the energy of moving object?

My question is simple. Work is measure of energy expended. If a moving object expends it's energy on a stationary object, wouldn't that stop it not the reaction force
The energy can go in many different ways. The point here is that your object is no longer an isolated object. It is interacting with other external force, i.e. the other object which provided another force or impulse. So not only should it not have the same energy afterwards, but Newton's Laws demand that it shouldn't be in the same kinematical situation.

Zz.

That step by step makes sense with that I was starting to think and understand. Only question though is why energy is define as the capacity for work. Clearly that seems very general when all that is going on.

Of course I guess many times when we deal with energy we would only look at the energy input and output on let's say the target object, which simplifies it alot

Here is a statement of the work-energy theorem: $$W_{AonB} = K_{B2} - K_{B1} = ΔK_{B}$$
This does not give us the exact energy change of A unless the interaction between A and B is conservative, i.e. elastic. In that case, by the conservation of mechanical energy we have $$K_{A1} + K_{B1} = K_{A2} + K_{B2}$$ and rearranging gives $$ΔK_{A} = -ΔK_{B}$$
So the energy change of one is the exact opposite of the energy change of the other. This explains the energy changes.

As far as why ball A stops, it is because of the law of conservation of momentum and this is founded upon empirical evidence (it is not dependent upon your being able to intuit this law).
$$m_{A1}v_{A1} + m_{B1}v_{B1} = m_{A2}v_{A2} + m_{B2}v_{B2}$$
When you do the math with simultaneous equations from the conservation of energy and the conservation of momentum in the case of elastic collisions you end up with ball A coming to a stop.

osilmag
sophiecentaur
Gold Member
That step by step makes sense with that I was starting to think and understand. Only question though is why energy is define as the capacity for work. Clearly that seems very general when all that is going on.

Of course I guess many times when we deal with energy we would only look at the energy input and output on let's say the target object, which simplifies it alot
You seem to have very high expectations of our past knowledge. Everything in Science is only 'near enough' and we always use the most coarse approximation we can get away with. The effects of all those Teratonnes of particles just on the limit of our observable Universe are too small to measure. So, apart from on a philosophical basis, why expect that to be brought into the Physics we are using?

1/r is a very powerful factor.

osilmag
Gold Member
From what you have described OP, the initial kinetic energy must equal the final kinetic energy. In an elastic collision, the momentum of object A would transfer totally to object B, stopping object A.

In an inelastic collision, there might also be friction in your case too which would mean the initial kinetic energy would equal the final kinetic energy plus the energy from friction. Not all of object A's momentum would be transferred to object B, so your conservation of momentum would still consider the final momentum of object A.

How is energy conserved with this?
This is how it works in the case of a moving object colliding with one at rest:
While the two objects are in contact
1. They are displaced by the same amount since they move together as one (that's what contact means).
2. The "projectile" object exerts a force in the same direction as the common displacement. Therefore, the projectile does positive work on the target.
3. The target object exerts an equal force but in the opposite direction (reaction counterpart) to the common displacement. Therefore, the target does negative work on the projectile.
4. The two works have the same magnitude because the forces have equal magnitudes at all times according to Newton's 3rd law.
5. The kinetic energy of the target increases because positive work is done on it while the kinetic energy of the projectile decreases because negative work is done on it.
Isn't the positive work creating energy in the target while the negative work is removing energy from the projectile? Energy is never created or destroyed right, but in this case it is kinda doing so within each objects, or atleast it appears that way.

Maybe I am reading into this too much. Perhaps it is just the way it is, as work is done, energy is increased/given to the target at the cost of the projectile having to be slow down/giving up energy.