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

  • Thread starter Thread starter Robertoalva
  • Start date Start date
  • Tags Tags
    Work Work done
Click For Summary
The discussion centers on the work done by the floor on a ball dropped from a height. When the ball rebounds, it reaches a height of only 5 m, indicating a loss of kinetic energy. The floor does exert a force during the collision, but since it does not move, it technically does no work in the traditional sense. The energy lost during the bounce is primarily due to the deformation of the ball and is converted into heat and sound, rather than being absorbed by the floor. The conversation highlights the complexities of energy transfer in inelastic collisions and the nuances of work done by forces.
  • #31
verty said:
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?
To be thorough, it may be that the floor deforms elastically while the ball deforms inelastically (or better, imperfectly elastically). In that case the floor does no net work. Or it could be the other way around, or anything in between.
 
Physics news on Phys.org
  • #32
darkxponent said:
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.
Instead of a uniform ball, consider a point mass attached atop a massless spring landing on the floor. In the frame of reference of an undeflected floor, the point mass does work on the spring, then, during rebound, the spring does work on the point mass. The spring is not moved by the floor, so the floor does no work.
You can extend this to a vertical chain of point masses joined by springs, thereby approximating a ball.
 
  • #33
Tanya Sharma said:
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 ?
Absolutely.

You invariably make complete sense.Thanks for the nice explanation.:smile:
You are most welcome. :wink:
 
  • #34
darkxponent said:
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
There's nothing wrong with applying dynamics to this problem. That's exactly what I did with the equation:
∫F.dx = ΔKE + ΔGPE

But you'll be missing something essential if you don't also consider energy. Since the floor does no real work on the ball, as I explained above, for the ball at rest to rise up from the floor there must be some stored elastic energy within the ball.

(I am assuming the floor doesn't move for this exercise. Imagine a rubber ball bouncing on a concrete floor.)
 

Similar threads

Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
8K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K