How does the floor contribute to the ball's energy during a bounce?

[Robertoalva]

1. A ball of mass 0.5 kg is dropped from rest 10 m above the floor. When it rebounds from the floor it comes to momentary rest 5 m above the floor. How much work did the floor do on the ball?
The acceleration due to gravity is 10 m/s2.

Homework Equations

W= F Δx

The Attempt at a Solution

so, I would say that, the work on the ball the first 10m is 50 J and then, it is half because the ball just went up 5 m and stopped so, the work done by the floor is -25J

 

[Robertoalva]

or is it that the floor doesn't have work?

 

[darkxponent]

1. A ball of mass 0.5 kg is dropped from rest 10 m above the floor. When it rebounds from the floor it comes to momentary rest 5 m above the floor. How much work did the floor do on the ball?
The acceleration due to gravity is 10 m/s2.

Homework Equations

W= F Δx

The Attempt at a Solution

so, I would say that, the work on the ball the first 10m is 50 J and then, it is half because the ball just went up 5 m and stopped so, the work done by the floor is -25J

That's correct.

or is it that the floor doesn't have work?

The floor does work at the time of collision.

 

[Doc Al]

A force does work if there is a displacement of the point of application of the force. Ask yourself: Does the floor move when it exerts its force? Can it do work?

 

[Robertoalva]

no, because if it doesn't move then F*0 = 0 ! then why is it that the correct answer is -25J?

 

[Doc Al]

no, because if it doesn't move then F*0 = 0 !

That's correct.

then why is it that the correct answer is -25J?

That's the change in kinetic energy, which is probably what they wanted. But it's not the work done by the floor.

Some books treat force * ΔX_cm as "work" (sometimes called center of mass work or pseudowork), where ΔX_cm is the displacement of the center of mass of the body. But that's not actual work done by the floor.

 

[rcgldr]

How much work did the floor do on the ball?

In real life, a floor compresses by a tiny amount and then recovers, so the floor does move a tiny amount of distance and can do work. In real life, most of the energy lost would be due to compression and recovery in the ball. Think of the portion of the ball that collides with the floor as a spring that presses downwards on the floor and upwards on the ball's center of mass. So I'm not sure how much of the work is done by the floor versus the work done by the ball, (unless the ball is an idealized ball that has totally elastic collisions (no energy lost).

 

[darkxponent]

That's correct.


That's the change in kinetic energy, which is probably what they wanted. But it's not the work done by the floor.

When there is change in kinetic energy there must be some Force which Works. Consider it just as a two body collision!

 

[Doc Al]

When there is change in kinetic energy there must be some Force which Works.

Not really. But there does have to be a force acting on it as the center of mass moves.

Consider it just as a two body collision!

Realize that the ball is a deformable body, not perfectly rigid. You cannot simply treat it as a point mass.

But rcgldr is correct that treating the floor as perfectly rigid is just an approximation.

 

[darkxponent]

In real life, a floor compresses by a tiny amount and then recovers, so the floor does move a tiny amount of distance and can do work. In real life, most of the energy lost would be due to compression and recovery in the ball. Think of the portion of the ball that collides with the floor as a spring that presses downwards on the floor and upwards on the ball's center of mass. So I'm not sure how much of the work is done by the floor versus the work done by the ball, (unless the ball is an idealized ball that has totally elastic collisions (no energy lost).

This is definitely an inelastic collision as total kinetic energy is not conserved. And the amount of kinetic energy lost by the ball will be lost as heat(this is what inelastic collision is). We can assume the Earth is does not move in the peoces of collision(as its mass is considered infinite). Just think of Earth applying some force on the ball, the ball deformes during the collision and the ball's centre of mass does move some some distance 'x' while Earth continues to apply the Force and hence the Earth does some Work on the ball.

Same force is applied to Earth by the ball, but as Earth's Centre of mass does not move, work done on Earth is zero.

 

[darkxponent]

Realize that the ball is a deformable body, not perfectly rigid. You cannot simply treat it as a point mass.

I din't assume ball to be rigid. Deformation does occur in inelastic collisions

 

[Doc Al]

TJust think of Earth applying some force on the ball, the ball deformes during the collision and the ball's centre of mass does move some some distance 'x' while Earth continues to apply the Force and hence the Earth does some Work on the ball.

Yes, the center of mass moves. But the point of application of the force (the floor) does not, thus that force technically does no work. (Yes, the floor not moving is just an approximation, but a reasonable one.)

 

[HallsofIvy]

The ball fell from 10 m, having gained kinetic energy of 10g J by the time it hit the floor. It only rebounded 5 me so had 5g J energy when it rebounded. It might be correct to say that the ball did 10g- 5g= 5g J work on the floor.

 

[Doc Al]

It might be correct to say that the ball did 10g- 5g= 5g J work on the floor.

Again, the floor doesn't move (to a good approximation) and thus no work is done on it.

Saying the ball does work on the floor, or that the floor does negative work on the ball, implies that the missing energy from the rebounding ball has gone into the earth. But no, most of that missing energy is in the compression/decompression/deformation of the ball and ends up as random thermal energy.

 

[darkxponent]

Yes, the center of mass moves. But the point of application of the force (the floor) does not, thus that force technically does no work. (Yes, the floor not moving is just an approximation, but a reasonable one.)

What do you mean by saying that it technically does no Work?

Yes, the center of mass moves. But the point of application of the force (the floor) does not,

Never heard of this "the movement of point of application of Force". What i knew was "Work is done when Force is applied to an abject and it moves some distance under the application of Force.". Do not understand by what you mean by "point of application of Force" or maybe i might not be familier to it.

 

[voko]

Saying the ball does work on the floor, or that the floor does negative work on the ball, implies that the missing energy from the rebounding ball has gone into the earth. But no, most of that missing energy is in the compression/decompression/deformation of the ball and ends up as random thermal energy.

This assumes a lot about the ball and the floor. Depending and what they really are, it may be entirely possible that most of the missing energy is in hysteretic losses in the floor, not the ball.

I think we can say that the problem is underspecified. Not knowing more about the ball and floor, one cannot say how much energy was dissipated by the floor and how much by the ball.

 

[verty]

If the wall does deflect but returns to its starting position, is there work done? Can we say that the ball does work deflecting the wall and the wall does work reversing that deflection?

 

[Doc Al]

What do you mean by saying that it technically does no Work?

No displacement = no work done.

Never heard of this "the movement of point of application of Force". What i knew was "Work is done when Force is applied to an abject and it moves some distance under the application of Force.". Do not understand by what you mean by "point of application of Force" or maybe i might not be familier to it.

Think of it this way: The floor exerts a force on the bottom of the ball, but the bottom of the ball does not move as long as contact (and the force) is maintained.

 

[Doc Al]

This assumes a lot about the ball and the floor. Depending and what they really are, it may be entirely possible that most of the missing energy is in hysteretic losses in the floor, not the ball.

I think we can say that the problem is underspecified. Not knowing more about the ball and floor, one cannot say how much energy was dissipated by the floor and how much by the ball.

All true.

 

[darkxponent]

Think of it this way: The floor exerts a force on the bottom of the ball, but the bottom of the ball does not move as long as contact (and the force) is maintained.

This is how i understand this collision process. The Force application is distributed so the point of application is moving. And those points are moving. During collision Force by ground is doing a negative work and at rebound it is doing Positive work, which is half of the negative work!

 

[Doc Al]

The Force application is distributed so the point of application is moving.

Not sure what you mean by that. The floor exerts a force on the part of the ball that is in contact with the floor. And, neglecting any motion of the floor itself, that part of the ball doesn't move during the collision.

Your diagram is fine, by the way.

 

[darkxponent]

Not sure what you mean by that. The floor exerts a force on the part of the ball that is in contact with the floor. And, neglecting any motion of the floor itself, that part of the ball doesn't move during the collision.

I think i got what you are saying. You agree that the Force is being applied on the body by the Ground.

∫F.dt = \frac{3}{2}mv

This force does no work but can change the momentum of the object. And this Force does move the center of mass of the object, changes the velocity of the Center of mass.

PS: It is the same way we jump, When we jump no work is done by the Earth on us?

 

[Doc Al]

I think i got what you are saying. You agree that the Force is being applied on the body by the Ground.

∫F.dt = \frac{3}{2}mv

This force does no work but can change the momentum of the object. And this Force does move the center of mass of the object, changes the velocity of the Center of mass.

Exactly! I'd also say that ∫F.dx = ΔKE, where dx is the displacement of the center of mass and KE is the KE of the center of mass.

PS: It is the same way we jump, When we jump no work is done by the Earth on us?

Exactly! (I was going to bring up this very example.) When you jump, your muscles are the source of the energy, not the earth.

 

[darkxponent]

Exactly! I'd also say that ∫F.dx = ΔKE, where dx is the displacement of the center of mass and KE is the KE of the center of mass.Exactly! (I was going to bring up this very example.) When you jump, your muscles are the source of the energy, not the earth.

It is very hard to understand that a Force acting on a object is affecting its momentum but not its Kinetic Energy. I know that the KE can be changed by internal Forces also, but change in momentum always requires some external Force.

This is very new concept to me 'that the point of application of the Force should move' otherwise the momentum will change and the change in KE will occur, but the reason behind the change in KE is some internal interaction, not that external Force!

 

[PhanthomJay]

The ball starts from rest and ends at rest at its rebound height. Since there is no KE change, the work energy theorem would appear to indicate that the work done by non conservative forces is equal to the GPE change, or - 25 J in this example. But the fact that no work is being done by the Normal non conservative external force contradict that, implying that you cannot always use this theorem when the mass is an object rather than a particle. It is perhaps best to look at conservation of energy, where in this example, the change in GPE plus the change in other forms of energy is 0, implying further that the change in other forms of energy must be 25 J, representing energy mostly in the form of heat and sound. This avoids the use of work.

 

[Tanya Sharma]

Some books treat force * ΔX_cm as "work" (sometimes called center of mass work or pseudowork), where ΔX_cm is the displacement of the center of mass of the body. But that's not actual work done by the floor.

Exactly! I'd also say that ∫F.dx = ΔKE, where dx is the displacement of the center of mass and KE is the KE of the center of mass.

Hi Doc

I am not clear about a few things...

If we apply work energy theorem ,

Work done by non conservative forces=Change in the mechanical energy of the ball

∫F.dx=ΔKE + ΔGPE + ΔEPE ,where ΔEPE is the change in the elastic potential energy of the ball and where dx is the displacement of the center of mass

Now ΔKE = 0
ΔGPE = -25J

Hence,ΔEPE = ∫F.dx + 25J

This ΔEPE is what goes away as thermal energy.

Do I make sense? If not where am I getting it wrong ?

 

[Tanya Sharma]

Doc Al , Is my reasoning correct in the above post ?

 

[Doc Al]

Hi Doc

I am not clear about a few things...

If we apply work energy theorem ,

Work done by non conservative forces=Change in the mechanical energy of the ball

Things get a bit tricky when dealing with deformable bodies such as this ball. In the so-called "work"-energy theorem, the "work" done by the external force of the floor will equal the change in mechanical energy. I put work in quotes, because it's really ∫F.dx where F is the force of the floor and x represents the motion of the center of mass. (So it's not the "real" work done by that force.)

∫F.dx=ΔKE + ΔGPE + ΔEPE ,where ΔEPE is the change in the elastic potential energy of the ball and where dx is the displacement of the center of mass

So, if you mean by ∫F.dx what I described above, then I would say that:
∫F.dx = ΔKE + ΔGPE
I would leave out the elastic PE of the ball, as that is internal. This equation is not particularly useful, since we have no simple way of calculating ∫F.dx.

On the other hand, using conservation of energy you can say:
Real Work done from floor = Δtotal energy
The real work is zero, so the total energy is conserved:
0 = ΔKE + ΔGPE + ΔEPE + other forms of internal energy, such as thermal energy. Now you can apply your analysis to that equation.

Now ΔKE = 0
ΔGPE = -25J

Using this, you'll get:
25 J = ΔEPE + other forms of internal energy, such as thermal energy.
When all is said and done, most likely all the mechanical energy lost ends up as internal thermal energy.

Hence,ΔEPE = ∫F.dx + 25J

This ΔEPE is what goes away as thermal energy.

Do I make sense? If not where am I getting it wrong ?

Let me know if my comments make sense.

 

[darkxponent]

Interna thermal energy you said Doc Al!.

What i am interested in is know what happens after the ball comes at rest(see my attachment).

During the collision, there is a time when ball is at rest. Now, what happens is that the ball starts moving in the opposite direction and finally comes to velocity v/2. That is momentum = mv/2 and K.E = mvv/8
I have to questions in mind.

1) What increases the momentum of the ball?- my answer would be the normal force by the ground. I came to this conclusion from Newtons second law.

2) What increases K.E of the ball?- My answer would be the Normal force, as it is responsible for increase in momentum so it is responsible for increase in KE.

Where am i wrong here? Is this not a mechanics question, rather a thermodynamics question. Do tell me

 

[Tanya Sharma]

Things get a bit tricky when dealing with deformable bodies such as this ball. In the so-called "work"-energy theorem, the "work" done by the external force of the floor will equal the change in mechanical energy. I put work in quotes, because it's really ∫F.dx where F is the force of the floor and x represents the motion of the center of mass. (So it's not the "real" work done by that force.)


So, if you mean by ∫F.dx what I described above, then I would say that:
∫F.dx = ΔKE + ΔGPE
I would leave out the elastic PE of the ball, as that is internal. This equation is not particularly useful, since we have no simple way of calculating ∫F.dx.

But that does give ∫F.dx = -25 J .Can we say that the center of mass work(not the "real" work done by that force) is equal to -25J ?

On the other hand, using conservation of energy you can say:
Real Work done from floor = Δtotal energy
The real work is zero, so the total energy is conserved:
0 = ΔKE + ΔGPE + ΔEPE + other forms of internal energy, such as thermal energy. Now you can apply your analysis to that equation.


Using this, you'll get:
25 J = ΔEPE + other forms of internal energy, such as thermal energy.
When all is said and done, most likely all the mechanical energy lost ends up as internal thermal energy.


Let me know if my comments make sense.

You invariably make complete sense.Thanks for the nice explanation.