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tncekm

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Okay, so I know that W=mgh=ΔUg. But, I'm a bit confused about how that relates to this equation:

W = ΔKE + ΔUg + ΔE

If you've got a block that travels vertical distance h down a frictionless incline θ you can "seemingly" use the above equation to find W, but it doesn't seem to work right.

There is no ΔE b/c there is no friction. So, you're left with:

W = ΔKE + ΔUg

From here, how do you get to W = ΔUg? There was a change in ΔKE, so I can't eliminate it!

Thanks!
 

iA-MD2013

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I never like using equations like the one you posted. It really does complicate things. We know that the only thing doing work on the object is gravity. W=mgh This work is used to accelerate the object and cause a change in Kinetic energy. From this alone, we can say that mgh=deltaKE.
Sorry I can't do better than this. Where did you get that equation from?
 

tncekm

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Its from EK.

That equation has helped clear some things up for me, but then it complicated the W=mgh equation. :(
 

wannabedocta

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I would only the the above equation using the displacement of the object, not the diagonal distance since work is independent of the path and only dependent on the initial and final positions. You might want to double check this, I may be wrong

W= F* displacement
or W= mgh
 

Kaustikos

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I would only the the above equation using the displacement of the object, not the diagonal distance since work is independent of the path and only dependent on the initial and final positions. You might want to double check this, I may be wrong

W= F* displacement
or W= mgh
wait, work is a state function? :confused:
 

tncekm

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Edit:

It looks like work can path independent when we're working with conservative forces.
 

RSAgator

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well...in this case W = delta_potential = delta_kinetic

I thought that EK's work/energy section was horrendous. You can solve 99% of all work problems using 3 forms of energy: heat, kinetic, and potential.

It's better to just consider initial and final states.

P = potential energy
K = kinetic energy
E = heat energy

Pi + Ki + Ei = Pf + Kf + Ef

or

(Pf - Pi) + (Kf - Ki) + (Ef - Ei) = d_P + d_K + d_E = 0

you can even be safe and say that

d_P + d_K + d_E = W

where W is nothing more than some form of energy that isn't accounted for in the equation. Work/energy is a pretty simple topic, just think of it as conservation. The energy in the system before has to be the energy in the system after.

In that EK equation the kinetic energy didn't drop off, I suspect it was simply absorbed into the generic W term. If you had a ball on top of a hill (Point A) and you let it roll to the bottom on a frictionless surface, then once at the bottom (point B) friction slowed it down to nothing (point C), then at point A W = 0, point B W = d_K = d_P, and point C W = d_P = d_E. I think more than anything it's just a term used to cover EK's bases so that it accounts for other sorts of atypical energy changes. Regardless, you're better off just comparing the final energy to the initial energy.
 

iA-MD2013

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well...in this case W = delta_potential = delta_kinetic

I thought that EK's work/energy section was horrendous. You can solve 99% of all work problems using 3 forms of energy: heat, kinetic, and potential.

It's better to just consider initial and final states.

P = potential energy
K = kinetic energy
E = heat energy

Pi + Ki + Ei = Pf + Kf + Ef

or

(Pf - Pi) + (Kf - Ki) + (Ef - Ei) = d_P + d_K + d_E = 0

you can even be safe and say that

d_P + d_K + d_E = W

where W is nothing more than some form of energy that isn't accounted for in the equation. Work/energy is a pretty simple topic, just think of it as conservation. The energy in the system before has to be the energy in the system after.

In that EK equation the kinetic energy didn't drop off, I suspect it was simply absorbed into the generic W term. If you had a ball on top of a hill (Point A) and you let it roll to the bottom on a frictionless surface, then once at the bottom (point B) friction slowed it down to nothing (point C), then at point A W = 0, point B W = d_K = d_P, and point C W = d_P = d_E. I think more than anything it's just a term used to cover EK's bases so that it accounts for other sorts of atypical energy changes. Regardless, you're better off just comparing the final energy to the initial energy.
I really hate the EK method too (just finished reading that section). The only way it works is if W=0, which is obviously not true. Work was done on the ball. But the work done was used to change it's kinetic energy...so I guess W=0 in the end. But I still hate that. I agree with dcohen, I think EK just does that to make you realize that work is done when there is a change in KE, when energy is lost to friction, and when there is change in potential. Use TPR or kaplan's methods. They are much more clear and straight forward.
 

tncekm

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Hmm... okie dokie. :D

So I look over the TPR method and I get the following:

W = ΔE = ΔUg + ΔK

That looks quite a bit different from W = ΔK + ΔUg + ΔE, where ΔE is added.

This was such a simple concept before I read over all of this stuff. :(
 

iA-MD2013

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Hmm... okie dokie. :D

So I look over the TPR method and I get the following:

W = ΔE = ΔUg + ΔK

That looks quite a bit different from W = ΔK + ΔUg + ΔE, where ΔE is added.

This was such a simple concept before I read over all of this stuff. :(
hahah yeah...I usually skim over these types of chapter where common sense is better than equations. Stick to TPR methods for physics.
 

RSAgator

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Hmm... okie dokie. :D

So I look over the TPR method and I get the following:

W = ΔE = ΔUg + ΔK

That looks quite a bit different from W = ΔK + ΔUg + ΔE, where ΔE is added.

This was such a simple concept before I read over all of this stuff. :(

See that equation isnt entirely accurate. If there's heat loss for instance, work won't equal the change in kinetic + the change in potential. You likely have a better intuitive understanding of the subject than the explanations given in the review books. The important concept is the conservation of energy. It's generally better to understand a concept than to memorize an equation, so just understand the following things:

energy can take many forms that are usually interconvertable between one another (though heat energy usually doesn't get converted back into "useful" energy that is, energy that can do work).

Energy is conserved - all energy present in the system at the initial state must be able to be accounted for at the final state (energy can neither be created nor destroyed).
 

tncekm

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Thanks for your replies.

I figured out the difference between
[1] W = ΔKE + ΔUg + ΔE (more appropriately W = ΔKE + ΔUg + ΔEi; no heat is present, but friction is, which can change Ei)
[2] W = ΔKE + ΔUg = ΔE (no heat is present, and no friction is present)

(1) should have had its E value stated as Ei, its internal energy and (2) is a case where there is no Ei. The E value in (2) is the "total energy" including PE And KE. So, that cleared things up a little bit.

I think I also figured out how to derive W=-ΔU

They're just "pretending" that all of the work goes into kinetic energy and there was no initial potential energy, and since K = -U, the equations is written as W=-mgh. Confusing, but I'll just deal with it.

*sigh* There are many hours of my life that I would love to get back after dealing with such a simple concept for so long today!
 

physics junkie

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This thread is giving me a headache.

Delta-Energy = Delta-Work + Delta-Heat. This equation always holds. Friction is energy lost to heat.

Starting Energy = Final Energy

For a closed system the starting energy is equal to the energy at the end because energy is conserved. So if you have a ball falling from a height h and it's at 0 velocity then it's starting energy term is mgh. If you want to know it's velocity at a height y you solve the equation mgh = mgy + (1/2)mv^2 for v. There are tons of ways you can formulate the starting and ending energy terms and it depends on the type of system you are looking at.

Another thing you might want to remember is that Total Energy = Kinetic energy + Potential Energy.

Most of the equations you've posted are special cases.
 
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