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

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SUMMARY

The discussion centers on the work done by the floor on a ball dropped from a height of 10 meters, which rebounds to a height of 5 meters. The ball, with a mass of 0.5 kg, experiences a gravitational force of 10 m/s², resulting in an initial kinetic energy of 50 J upon impact. The work done by the floor is calculated as -25 J, indicating energy loss primarily due to inelastic collision and deformation of the ball, rather than work done by the floor itself, which does not move during the collision.

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  • #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.
 
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  • #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.)
 

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