Physics Help Forum Qick Thermodynamics Quiz

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Mar 7th 2018, 08:02 AM #1 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 820 Qick Thermodynamics Quiz A small block is placed in a smooth bowl and held partway up the side of the bowl and then released, whereupon it slides down towards the bottom. Discuss the system and How much work is done?
 Mar 7th 2018, 03:09 PM #2 Senior Member   Join Date: Jun 2010 Location: NC Posts: 377 A subtle situation, you propose The physical situation you have posed and requested analysis, has been on my mind quite some time (interestingly, near precisely as you posed it). This relevant physics history I don't know. Me bing engineering educated, your mention of "system" is comforting. The issue about "work" here (again I don't know) seems to me definitional. (same word - two definitions). Newton's 2nd, the force sum, had surface and body forces in it. The idea "work" happened when the 2nd, sum of forces were made product with displacement. A split in physics thinking happened. It made sense: "Surface force-displacement was WORK." Gravity force displacement was a scalar = potential energy change. (How did this come to pass?) So, to my thinking, everytime any physics person mentions PE; they have that something AND Earth as the system. I've written about the same in my notes (as above) often. I believe what I say happened. My resource texts don't say. So, good luck with your thread. JP studiot likes this. Last edited by THERMO Spoken Here; Mar 7th 2018 at 03:11 PM.
 Mar 8th 2018, 12:16 AM #3 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 820 Thermo, thank you for your thoughts. You have indeed identified a possible misunderstanding. In fact this question was prompted by the two recent threads about common misconceptions in Physics and I was actually unsure whether to post it there or start a new thread. In the end, since the answer lies in Thermodynamics I started a new one here in the thermo section.
Mar 8th 2018, 06:54 AM   #4
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 Originally Posted by studiot A small block is placed in a smooth bowl and held partway up the side of the bowl and then released, whereupon it slides down towards the bottom. Discuss the system and How much work is done?
I'm not sure thermodynamics is relevant here...

Since there is no friction, the amount of gravitational potential energy converted to heat and sound is negligibly small, so straightforward GPE -> KE conversions should do it.

Note that if there is no friction, the block should oscillate up and down the bowl, with the height gained on the other side being the same as the initial height (or at least very close to it if you want to consider the tiny heat and sound loss).

As for work? GPE and KE will keep converting between each other, so work will continuously be done on the block by the net residual force (gravitational field + the pushing force of the Earth+bowl system) until someone comes along and tries to stop it. However, when the block is swinging back from the other side of the bowl back to the original side, the work done becomes negative (because of path-independence of the work-energy principle), so if you want to obtain the total amount of energy exchange occurring (over some duration of interest), you'll need to take the absolute value of F(x) and integrate that instead.

Mar 8th 2018, 07:55 AM   #5
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 Originally Posted by benit13 I'm not sure thermodynamics is relevant here... Since there is no friction, the amount of gravitational potential energy converted to heat and sound is negligibly small, so straightforward GPE -> KE conversions should do it. Note that if there is no friction, the block should oscillate up and down the bowl, with the height gained on the other side being the same as the initial height (or at least very close to it if you want to consider the tiny heat and sound loss). As for work? GPE and KE will keep converting between each other, so work will continuously be done on the block by the net residual force (gravitational field + the pushing force of the Earth+bowl system) until someone comes along and tries to stop it. However, when the block is swinging back from the other side of the bowl back to the original side, the work done becomes negative (because of path-independence of the work-energy principle), so if you want to obtain the total amount of energy exchange occurring (over some duration of interest), you'll need to take the absolute value of F(x) and integrate that instead.
Can you put that into the first law?

That is what is the change in internal energy of the block?

Last edited by studiot; Mar 8th 2018 at 09:12 AM.

Mar 8th 2018, 06:32 PM   #6
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 Originally Posted by benit13 I'm not sure thermodynamics is relevant here....

 Originally Posted by benit13 Since there is no friction,.
No friction ??? where does it say that , smooth does not mean frictionless !!

Depending on how much wind resistance and friction there is , it may go up the other side and back again etc ...or may not.

Mar 9th 2018, 12:20 AM   #7
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 No friction ??? where does it say that , smooth does not mean frictionless !! Depending on how much wind resistance and friction there is , it may go up the other side and back again etc ...or may not.
Smooth is standard shorthand for no friction or other resistance forces acting.
So q = 0

OK so the first law says

dU = q + w (depending upon your sign convention)

and dU = 0

so w = 0

 Mar 9th 2018, 02:46 AM #8 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 112 The work done goes into kinetic energy of the block, not the internal energy, so $\displaystyle \Delta Q = 0, \Delta W = 0$ Therefore $\displaystyle \Delta U = 0$ It's not very fun though... why not try adding a perturbation to the net force and see how it propagates to the work done and therefore the internal energy? You could even set up an equation associated with the thermal equilibrium of the block with the surroundings
Mar 9th 2018, 03:42 AM   #9
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 Originally Posted by benit13 The work done goes into kinetic energy of the block, not the internal energy, so $\displaystyle \Delta Q = 0, \Delta W = 0$ Therefore $\displaystyle \Delta U = 0$ It's not very fun though... why not try adding a perturbation to the net force and see how it propagates to the work done and therefore the internal energy? You could even set up an equation associated with the thermal equilibrium of the block with the surroundings
So are you saying the internal energy of the block doesn't change, even though it's potential energy changes?

Mar 12th 2018, 03:01 AM   #10
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 Originally Posted by studiot So are you saying the internal energy of the block doesn't change, even though it's potential energy changes?
Yes. Since the block doesn't change its temperature, it doesn't increase its internal energy.

But, like I suggested, there's usually a tiny bit of energy going into heat and sound, even with surfaces with low coefficients of friction. That's why it would be more fun to include it

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