What is my textbook trying to say?

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The discussion revolves around the application of energy conservation principles to a problem involving a box sliding down an incline with friction. The initial calculations presented yield different speeds for the box, leading to confusion over the correct application of potential energy, kinetic energy, and work done by friction. Participants emphasize the importance of consistent reference points for potential energy and displacement, noting that the zero point for height and displacement can differ. The conversation highlights the necessity of careful sign management in equations involving non-conservative forces like friction. Ultimately, the participants agree that as long as the conventions are consistently applied, the results should align, regardless of the chosen reference points.
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Homework Statement



[PLAIN]http://img830.imageshack.us/img830/3263/19000013.jpg A 12kg box initially at rest slides from the top of an incline. If the force of friction along the incline is 5.0N, what is the speed of the object as it reaches the bottom of the incline?

My book is contradicting itself?

Textbook said:
ΔKE + ΔPE + ΔTE = 0

ΔKE + ΔPE = -ΔTE

½m(v² - v₀²) + mg(h - h₀) = -Ffrx

½(12kg)(v² - 0) + (12kg)(9.8m/s2)(0 - 6.0m) = -(5.0N)(11.0m)

v = \sqrt{\frac{651J}{6.0kg}}

v = 10m/s
Here is what I think it should be

ΔKE + ΔPE + ΔTE = 0

½m(v² - v₀²) + mg(h - h₀) + Ffr(x - x₀) = 0

½(12kg)(v² - 0) + 12(9.8)(0 - 6) + (5.0N)(0 - 11m) = 0

6v² - 705.6J - 55J = 0

v = \sqrt{\frac{760J}{6kg}} = 11.25m/s
 
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In the expression for friction, x = 11 and x0= 0 as this is the initial displacement. Initially, at time = 0, the block does not move so there is no initial displacement.
 
rock.freak667 said:
In the expression for friction, x = 11 and x0= 0 as this is the initial displacement. Initially, at time = 0, the block does not move so there is no initial displacement.

But I thought the question set it up so that the bottom is 0, that's how it can be consistent with the change in potential energy I had before, otherwise I had to change that too.
 
flyingpig said:
But I thought the question set it up so that the bottom is 0, that's how it can be consistent with the change in potential energy I had before, otherwise I had to change that too.

The zero line for the height is not the same as the zero line for the 'x'.
 
rock.freak667 said:
The zero line for the height is not the same as the zero line for the 'x'.

But how do you know and why isn't it the same?
 
Here is another question where they used the my convention

problem said:
A 2.5kg box slides from rest 0.850m down a 30 degree incline. If the force of friction acting along the incline is 3.2N, what is the speed of the box when it reaches the bottom of the incline?

Textbook solution said:
Δh =0 - 0.850sin30 = -0.425m

ΔKE + ΔPE + ΔTE = 0

ΔKE + ΔPE = -ΔTE

½m(v² - v₀²) + mgΔh = -Ff • d

½(2.5kg)(v² - 0²) + (2.5kg)(9.80m/s^2)(-0.425m) = -(3.2N)(0.850m)

1.25v² = -2.72J + 10.42J

v = 2.5m/s

Which is the way I originally did it, but this method contradicts the first question completely.
 
In case of non-conservative force(s), I prefer to use the Work-Energy Theorem.
The change of kinetic energy is equal to the sum of work of all forces.

ΔKE=W(gravity) + W(friction).

Work is force times displacement in the direction of the force.

The box moves from the top to the bottom of the incline. In case the origin is at the initial position of the box, and consider the downward directions positive, the vertical force is mg, and the vertical displacement is h. The displacement along the slope is L=11 m. The force of friction is opposite to the motion so the work of friction is -L*Fr (with Ffr=50 N).

So ΔKE=mgh - LFfr.

In terms of conservation of energy, you need to add the negative of the work of friction to the change of the mechanical energy, in order to make the sum zero.

ΔKE+ΔPE-Wfr=0

The origin is at the initial position:

KE(initial)=0, KE (final)=1/2 mv2--->

ΔKE=1/2 mv2

You can choose the zero of potential energy at any point. If you choose it zero at the origin :

PE(initial)=0, PE(final)=-mgh--->ΔPE=-mgh

Th ework of friction is

Wfr=-Ffr*L.

All of these result in

1/2 mv2-mgh-(-FfrL)=0--->

1/2 mv2=mgh-L Ffr.



ehild
 
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The box moves from the top to the bottom of the incline. In case the origin is at the initial position of the box

No, the initial is at the top not at the origin.

You can choose the zero of potential energy at any point. If you choose it zero at the origin

But I am trying to be consistent with the triangle
 
I assume ΔTE stands for the change in thermal energy. Friction will cause an increase in thermal energy, and since the work done by friction Wfr is negative, you must have ΔTE=-Wfr.

There are a couple of ways you can calculate the work done by friction. Since the force of friction is constant in this problem, we can use

W = \mathbf{F}\cdot \Delta\mathbf{x}

One way is to use the fact that \mathbf{A}\cdot\mathbf{B}=|A||B|\cos\theta, where θ is the angle between the two vectors. The force of friction opposes the motion, so θ=180 degrees, so we get

W = |\mathbf{F}||\Delta\mathbf{x}|\cos 180^\circ = -(5~\mathrm{N})(11~\mathrm{m}) = -55~\mathrm{J}.

Another way is to use the fact that \mathbf{A}\cdot\mathbf{B}=A_x B_x + A_y B_y + A_z B_z. Let's use your convention where the block starts at x0=11 m and ends at x=0 m. That means the positive direction points up the incline, and Δx = -11i m. Because the force of friction points up the incline, we have F = +5i N, so the work is equal to

W = (5\hat{i}~\mathrm{N})\cdot(-11\hat{i}~\mathrm{m}) = (5~\mathrm{N})(-11~\mathrm{m}) = -55~\mathrm{J}

If instead we used the convention that x0=0 m at the top of the incline and x=11 m at the bottom, the positive direction is down the incline. Therefore, the force of friction would be F = -5i N, and the displacement would be Δx = 11i m. The work would then be

W = (-5\hat{i}~\mathrm{N})\cdot(11\hat{i}~\mathrm{m}) = (-5~\mathrm{N})(11~\mathrm{m}) = -55~\mathrm{J}

You get the same answer in each case, as you should. It doesn't matter which convention you follow as long you're consistent.
 
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  • #10
flyingpig said:
No, the initial is at the top not at the origin.
But I am trying to be consistent with the triangle

You can place the origin anywhere, the triangle will not change. Why not put the origin of your coordinate system at the initial position of the box, which is at the top of the incline? It does not matter, as Vela explained.

ehild
 
  • #11
No, the point is, in that equation, if you are trying to be consistent with your triangle, the signs will switch
 
  • #12
How so? Show your calculations and specifically show how the sign switches.
 
  • #13
Just the first problemΔKE + ΔPE + ΔTE = 0

ΔKE + ΔPE = -ΔTE

½m(v² - v₀²) + mg(h - h₀) = -Ffrx

½(12kg)(v² - 0) + (12kg)(9.8m/s2)(0 - 6.0m) = -(5.0N)(11.0m) <==== at here it should be -(5.0N)(-11.0m) = 55N

If I used the dot product in the vectorial form as you did in the second method, I should get a positive because you said ΔTE=-Wfr.

Which is consistent with what I am doing ΔTE=-(-55N) = 55N
 
  • #14
I explained in post #9 why the signs work out the way they do. You get the same answer regardless of where you place the origin x0 and which way you orient the x axis.
 
  • #15
flyingpig said:
Just the first problem


ΔKE + ΔPE + ΔTE = 0

ΔKE + ΔPE = -ΔTE

½m(v² - v₀²) + mg(h - h₀) = -Ffrx

½(12kg)(v² - 0) + (12kg)(9.8m/s2)(0 - 6.0m) = -(5.0N)(11.0m) <==== at here it should be -(5.0N)(-11.0m) = 55N
Are you saying the RHS should be +55 J and not -55 J? Remember ΔTE is positive because friction is turning the mechanical energy in the system into heat, so -ΔTE is negative.
If I used the dot product in the vectorial form as you did in the second method, I should get a positive because you said ΔTE=-Wfr.

Which is consistent with what I am doing ΔTE=-(-55N) = 55N
 
  • #16
½m(v² - v₀²) + mg(h - h₀) + µmg(x - x₀) = 0

Is this correct? I'll come back to the dot product later since this version is almost the same

Because i think my problem lies there
 
  • #17
Depends. What do you think x and x0 equal to?
 
  • #18
It doesn't matter, x is the final position and x₀ is the initial position consistent with h and h₀ in mg, ALWAYS!
 
  • #19
Well, the problem is you can't neglect the vector nature of the quantities in the dot product. The expression you wrote only works if x-x0>0. I already explained this back in post 9, so I won't bother repeating it.

Another point people have implied that I think you're missing is that the convention you use for h is independent from the convention you use for x. You could perfectly well have h=0 at the top of the incline and x=0 at the bottom. As long as you're consistent, everything will work out.
 
  • #20
Sorry for the late response, I had so much to do over the break!

But if you could "set it anyway" you want, then how do you know which one is correct? Why does it work when it is going up and it doesn't work when it is going down?
 
  • #21
Go back and read post #9. I went over the different cases there. In particular, look at the last two examples.
 
  • #22
What I did said:
½(12kg)(v² - 0) + 12(9.8)(0 - 6) + W = (5\hat{i}~\mathrm{N})\cdot(-11\hat{i}~\mathrm{m}) = (5~\mathrm{N})(-11~\mathrm{m}) = -55~\mathrm{J} = 0

Textbook said:
½(12kg)(v² - 0) + (12kg)(9.8m/s2)(0 - 6.0m) = -(5.0N)(11.0m)

I used your method and it still disagrees. I don't understand
 
  • #23
How did you derive this equation?

½(12kg)(v² - 0) + 12(9.8)(0 - 6) + W = 0
 
  • #24
It's just this ΔKE + ΔPE + ΔTE = 0
 
  • #25
Reread the first paragraph of post 9.
 
  • #26
Wait, is that because friction is a non-conservative force?

But why doesn't it work in this problem?

problem said:
A constant force 85N accelerates a 12kg box from a speed of 2.0m/s to a speed of 8.0m/s as it travels 15m along a horizontal surface. What is the coefficient of friction between the two surfaces?

I did (using your method)

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

So why doesn't it work on this problem?
 
  • #27
You need to keep track of the signs much more carefully. You tend to add and drop signs here and there willy-nilly which is why nothing ever works out.
 
  • #28
What do you mean? I followed it exactly!
 
  • #29
No, you didn't. You really didn't.
 
  • #30
vela said:
No, you didn't. You really didn't.


I assume ΔTE stands for the change in thermal energy. friction will cause an increase in thermal energy, and since the work done by friction Wfr is negative, you must have ΔTE=-Wfr.

Isn't that what I did too?
 
  • #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.
 
  • #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:
 

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