Mechanical energy of ball in water

[SteelDirigibl]

Homework Statement

A 2.00 kg ball, initially traveling at 10.0m/s, falls straight down through 10.0m of water before (momentarily) stopping. By how much does its mechanical energy change?

Homework Equations

1/2*mv²

The Attempt at a Solution

I thought I could just use the KE equation, but apparently that is wrong. I got -100J, plugging those numbers into the equation. Seemed fairly straightforward. But I didn't use the 10.0m displacement through water, or anything to account for the fact it traveled through water. I thought it wouldn't matter. According to my answer sheet, the answer is -296J. I'm not sure how to get this, and can't find the relevant section in my book.

So how does the water affect the change in energy, and how do I get to -296 Joules?
This is a practice final for my physics class, and I have the answer key, so that is how I know it is -296J, I just can't get there. If it's something with fluid dynamics, I'm pretty sure we didn't do any of that, so perhaps it's just not something we learned. Regardless, I would like to know how to do it, because it seems fairly simple.

 

[Precursor]

Mechanical energy considers both kinetic and potential energy.

 

[SteelDirigibl]

ah... right...

so I would add my -100J that I found to the PE at the beginning... 2.00kg*10.0m*9.8=196J

but it's all negative because it has no energy at the instant it hits the bottom? 0 potential and 0 kinetic.

-100J-196J=-296J

 

[Precursor]

That's correct. The ball loses both kinetic and potential energy.

 

[stevenb]

That appears to be the answer they are looking for, but I'm confused about something.

The problem says that the ball stops momentarily. Doesn't this imply that the ball is buoyant? In other words, if it were denser than water, wouldn't it keep sinking and not stop till it hits bottom?

If it is buoyant, then hasn't it gained additional unaccounted potential energy by doing work against buoyant force? Or, another way to think of it is that the gravitational potential energy is different than mgh since the force of gravity would not be felt underwater ( even if the ball is denser than water, then full force of gravity is not felt).

It's clear they don't give enough information to quantify the buoyant force, but I find this a misleading question. - Or, am I overlooking something simple?

 

[SteelDirigibl]

perhaps it means it stops momentarily before it bounces back up? And is saying just that it is a 10.0m deep container. Of course, I don't see the point of even mentioning the fact that it goes through water then. Perhaps only to imply that it wouldn't bounce? (some energy is lost due to resistance to water, though it would still lose the same amount?)

At any rate, unless I completely missed something, we didn't do anything related to the fluid dynamics aspect this semester. I'm not sure why water was even mentioned.

 

[stevenb]

perhaps it means it stops momentarily before it bounces back up? And is saying just that it is a 10.0m deep container. Of course, I don't see the point of even mentioning the fact that it goes through water then. Perhaps only to imply that it wouldn't bounce? (some energy is lost due to resistance to water, though it would still lose the same amount?)

At any rate, unless I completely missed something, we didn't do anything related to the fluid dynamics aspect this semester. I'm not sure why water was even mentioned.

I think you hit the reason for the water. Basically, it seems to be there to dissipate the kinetic energy without allowing the energy to be stored elastically and recovered in a bounce. But, what bothers me is that this then makes gravitational potential energy ambiguous, if you are not given more information.

Of course, leaving out the information that would allow you to compute the buoyant force makes it clear that they want you to ignore that aspect. And, the final answer makes that crystal clear.

I'm probably being too nitpicky, or maybe just missing a simple point, but it might be something worth mentioning to your teacher.

 

[PhanthomJay]

That appears to be the answer they are looking for, but I'm confused about something.

The problem says that the ball stops momentarily. Doesn't this imply that the ball is buoyant?

yes

In other words, if it were denser than water, wouldn't it keep sinking and not stop till it hits bottom?

Under certain conditions in accord with Archimeded Principle, it'd keep on going...

If it is buoyant, then hasn't it gained additional unaccounted potential energy by doing work against buoyant force?

By doing work against the bouyant force, it loses its initial mechanical energy it had at the surface

Or, another way to think of it is that the gravitational potential energy is different than mgh since the force of gravity would not be felt underwater ( even if the ball is denser than water, then full force of gravity is not felt).

Gravitational potential energy at a certain point is always mgh, where h is the vertical distance from a reference point; What is felt underwater is the bouyant force.

It's clear they don't give enough information to quantify the buoyant force, but I find this a misleading question. - Or, am I overlooking something simple?

The change in the objects mechanical energy is due to the work done by the buoyant force.

 

[SteelDirigibl]

so then the question is accurate? basically we are finding the work done by the buoyant force.

but even if there were no buoyant force, say it just dropped of a ladder 10m high, and the ball WERE to bounce back off a hard surface, at that instant momentary stop at the bottom of the 10m, the change in energy would have gotten the same answer. right? meaning the use of water still isn't necessary, unless it was just intended to make me think about the buoyancy.

 

[PhanthomJay]

so then the question is accurate? basically we are finding the work done by the buoyant force.

but even if there were no buoyant force, say it just dropped of a ladder 10m high, and the ball WERE to bounce back off a hard surface, at that instant momentary stop at the bottom of the 10m, the change in energy would have gotten the same answer. right? meaning the use of water still isn't necessary, unless it was just intended to make me think about the buoyancy.

If it bounced off a very hard surface it wouldn't stop instantly at 10.000 meters, it would have to deform and stop in a distance slightly greater. The problem could have said I suppose that the ball fell into a pile of ultra soft loosely packed styrofoam pellets and came to a stop in 10 m, and the answer would be the same...just less wet ...and with no bounce

 

[stevenb]

The change in the objects mechanical energy is due to the work done by the buoyant force.

Yes, I get that. The initial kinetic and potential energy the ball has above the water surface gets taken up by frictional dissipation in the water and by work done against buoyant force.

The dissipated energy is lost and the ball no longer has that mechanical energy.

However, isn't the potential energy of a submerged buoyant ball also mechanical energy since the ball can now do useful work as it surfaces again? And, if so, doesn't the ball still have this mechanical energy, and shouldn't that amount be subtracted from the answer of 296 J?

Or, maybe, it's better to say that the water has the potential energy and then gives it back to the ball later? Maybe that's what I'm missing.

 

[PhanthomJay]

Yes, I get that. The initial kinetic and potential energy the ball has above the water surface gets taken up by frictional dissipation in the water and by work done against buoyant force.

The dissipated energy is lost and the ball no longer has that mechanical energy.

However, isn't the potential energy of a submerged buoyant ball also mechanical energy since the ball can now do useful work as it surfaces again? And, if so, doesn't the ball still have this mechanical energy, and shouldn't that amount be subtracted from the answer of 296 J?

At the bottom of the fall, the ball has neither potential or kinetic energy...it has no ability to do useful work on its own...the useful work done is provided by the non-conservative buoyant force, pushing it back up.

 

[stevenb]

At the bottom of the fall, the ball has neither potential or kinetic energy...it has no ability to do useful work on its own...the useful work done is provided by the non-conservative buoyant force, pushing it back up.

Yes, i see it now. Thanks!

 

[stevenb]

the useful work done is provided by the non-conservative buoyant force, pushing it back up.

By the way, what is your basis for considering the buoyant force to be non-conservative? It seems conservative to me.

 

[Precursor]

The buoyant force is essentially friction. Therefore, the ball loses its mechanical energy due to friction.

 

[stevenb]

The buoyant force is essentially friction.

Buoyant force is different than friction force. Do you mean "essentially" from the point of view that you can have a situation where buoyant force and friction force are equal in magnitude and opposite in direction? That can happen if the ball is allowed to rise due to buoyant force and then reaches a terminal velocity due to friction balancing that force so that there is no further acceleration.

This also relates to my question of why buoyant force would be non-conservative. I would think buoyant force is conservative, but we know that friction forces are not generally conservative.

 

[SteelDirigibl]

i think essentially they are the same in the way they affect the motion of the ball. both are a force that opposes the direction of motion. energy would be lost from the work done in moving the water however, where in friction energy is lost in heat/sound etc. The overall affect as a whole would be the same though.

 

[stevenb]

i think essentially they are the same in the way they affect the motion of the ball. both are a force that opposes the direction of motion. energy would be lost from the work done in moving the water however, where in friction energy is lost in heat/sound etc. The overall affect as a whole would be the same though.

Yes, that makes sense.