What is my textbook trying to say?

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Homework Help Overview

The discussion revolves around a physics problem involving a box sliding down an incline, with a focus on the application of energy conservation principles and the work-energy theorem. Participants are examining the effects of friction and the definitions of potential and kinetic energy in the context of the problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore different formulations of energy conservation equations, questioning the setup of initial and final positions for both height and displacement. There are discussions about the signs used in equations and the implications of friction on energy transformations.

Discussion Status

Multiple interpretations of the problem are being explored, particularly regarding the conventions for defining potential energy and displacement. Some participants suggest that the choice of origin for height and displacement can lead to different signs in equations, while others emphasize consistency in applying these conventions.

Contextual Notes

There is an ongoing debate about the definitions of variables and the assumptions made in the problem setup, particularly concerning the initial conditions and the nature of the forces involved. The discussion reflects a lack of consensus on how to approach the problem while adhering to the principles of physics.

  • #31
I don't know. Maybe you did, maybe you didn't. We don't know since you haven't shown any intermediate steps. All I know is your final equation is wrong.
 
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  • #32
vela said:
I don't know. Maybe you did, maybe you didn't. We don't know since you haven't shown any intermediate steps. All I know is your final equation is wrong.


ΔKE + ΔPE + ΔTE = 0

ΔKE + ΔPE - W = 0

½m(v² - v₀²) + mg(h - h₀) - Ffr(x - x₀) = F • (x - x₀)

½(12)(8² - 2²) + 0 - µ(12)(9.8)(15 - 0) = 85N • (15 - 0)

µ = -0.51


^Intermediate steps!
 
  • #33
So you're saying F • (x - x₀)=0 and W=Ffr(x - x₀)?
 
  • #34
No

½m(v² - v₀²) + mg(h - h₀) - Ffr(x - x₀) = F • (x - x₀)

but W is Ffr(x - x₀) = -ΔTE
 
  • #35
Again, go back and read post 9 on how to calculate the work correctly.
 
  • #36
I still don't understand, the method looks the same, but it doesn't work
 
  • #37
Please I really don't understand. I read your paragraph over and over again and I applied, but it doesn't work!
 
  • #38
flyingpig said:
Please I really don't understand. I read your paragraph over and over again and I applied, but it doesn't work!

Okay, for this second problem that you posted on page 2 of the thread, we can just used the work-energy theorem directly. The work done on the object is equal to its change in kinetic energy. There is not much point in including a potential energy term, since no conservative forces are at play.

W = ΔKE

Now, the work done can be divided into the work done by the applied (85 N) force, and the work done by the frictional force.

Wapp + Wfric = ΔKE

Or:

ΔKE - Wfric = Wapp

Now, at this point, I want to emphasize that the W's each have their own intrinsic signs. Although Wfric has a negative sign in front of it in the equation above, it is also intrinsically negative. The question is, how to arrive at this result, and be sure that you've got the signs right? In post #9, vela gave you an extremely detailed account of how to do so. The thing to remember is that the work done is the dot product of the force vector and the displacement vector. In what follows, I'll use boldface quantities to represent vectors.

ΔKE - Ffric • Δx = Fapp • Δx

How to calculate the dot products? You've been given two methods in this thread. One of them is to use the result that the dot product of two vectors is the product of the magnitudes of those two vectors, multiplied by the cosine of the angle between them. Since the frictional force always opposes the motion, the angle between it and the displacement is 180 degrees. In contrast, the applied force is in the direction of motion, meaning that it makes a zero degree angle with the displacement vector:

ΔKE - |Ffric||Δx|cos(180°) = |Fapp||Δx|cos(0°)

Evaluate the cosines:

ΔKE - |Ffric||Δx|(-1) = |Fapp||Δx|(1)

Combine all the negative signs and get rid of the factors of 1:ΔKE + |Ffric||Δx| = |Fapp||Δx|

Now, at this point, to make things easier to read, if you want you can use non-boldface symbols to represent the magnitudes of the vectors that appear above. I.e. F = |F| and Δx = |Δx|. The reason I didn't do this before is because I wanted to make it clear that THESE are the careful steps with the vectors that you failed to carry out, and that is why you never got the signs right.ΔKE + FfricΔx = FappΔx

(1/2)m(v2 - v02) + µmg(x - x0) = Fapp(x - x0)

Now, I think you'll find that if you plug in numbers, things will work out fine.
 
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  • #40
hi flyingpig! :wink:
flyingpig said:
ΔKE + ΔPE + ΔTE = 0

6v² - 705.6J - 55J = 0

∆TE (the increase in thermal energy) is minus the work done by friction (just as ∆PE is minus the work done by gravity) …

work done is force "dot" displacement, and since friction is always opposite to the displacement, the work done is negative, and so ∆TE is positive :smile:

(by comparison, in this case gravity is in approximately the same direction as the displacement so the work done is positive, and so ∆PE is negative)
 
  • #41
cepheid said:
Okay, for this second problem that you posted on page 2 of the thread, we can just used the work-energy theorem directly. The work done on the object is equal to its change in kinetic energy. There is not much point in including a potential energy term, since no conservative forces are at play.

W = ΔKE

That is the work-energy theorem? I am trying to use conservation of energy
 
  • #42
flyingpig said:
That is the work-energy theorem? I am trying to use conservation of energy

Yes, the work-energy theorem says that the net work done on an object is equal to its change in kinetic energy.

Use conservation of energy if you want. You'll end up with exactly the same equation that I did.
 
  • #43
cepheid said:
Yes, the work-energy theorem says that the net work done on an object is equal to its change in kinetic energy.

Use conservation of energy if you want. You'll end up with exactly the same equation that I did.

No that is the nature of this problem, I used the conservation of energy, but it didn't work
 
  • #44
flyingpig said:
No that is the nature of this problem, I used the conservation of energy, but it didn't work

Yeah, but even if you use "two different methods", you should arrive at the same answer, right? (I put that phrase in quotation marks, because they are really just two different ways of looking at the same thing).

Conservation of energy says: the TOTAL* energy you put into the system (which is just the applied work) is equal to the change in energy of the system (taking into account all forms of energy that are present). If this were not true, then energy would be disappearing or coming from nowhere. There is no change in potential energy. So the only types of energy that change are thermal and kinetic:

ΔKE + ΔTE = Wapp (cons. of energy)

Now, as tiny-tim explained above, the change in thermal energy is the negative of the work done by friction, so that we get:

ΔKE -Wfric = Wapp

Notice that this equation is now EXACTLY THE SAME as the one I came up with in my previous post (using slightly different words to describe my method). Therefore, all of the rest of the steps I carried out after this one in my previous post still apply. Read them carefully. PLEASE.

*as opposed to the NET work done, which is the part that becomes kinetic.
 
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  • #45
Can I say then the change in thermal energy equals the negative work done by friction just as the change in PE is the negative work done by gravity?

ΔKE = ∑W_{work done by all other forces}
 
  • #46
in this case, yes (∆TE = Ffr.d) …

because the only thing reducing the mechanical energy (KE + PEgrav) in this case is the friction …

but if for example the block collided with something, there would (probably) be no "conservation of energy" (ie mechanical energy), and the "lost energy" becomes the gained thermal energy. and that of course comes not from friction but from the stresses inside the material during the collision :wink: