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What is my textbook trying to say?

  • Thread starter flyingpig
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  • #1
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Homework Statement



[PLAIN]http://img830.imageshack.us/img830/3263/19000013.jpg [Broken]


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 = [tex]\sqrt{\frac{651J}{6.0kg}}[/tex]

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 = [tex]\sqrt{\frac{760J}{6kg}}[/tex] = 11.25m/s
 
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Answers and Replies

  • #2
rock.freak667
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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.
 
  • #3
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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.
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.
 
  • #4
rock.freak667
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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'.
 
  • #5
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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?
 
  • #6
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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.
 
  • #7
ehild
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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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  • #8
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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
 
  • #9
vela
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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

[tex]W = \mathbf{F}\cdot \Delta\mathbf{x}[/tex]

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

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

Another way is to use the fact that [itex]\mathbf{A}\cdot\mathbf{B}=A_x B_x + A_y B_y + A_z B_z[/itex]. 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

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

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

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

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
ehild
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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
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No, the point is, in that equation, if you are trying to be consistent with your triangle, the signs will switch
 
  • #12
vela
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How so? Show your calculations and specifically show how the sign switches.
 
  • #13
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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

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
vela
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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
vela
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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
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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
vela
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Depends. What do you think x and x0 equal to?
 
  • #18
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It doesn't matter, x is the final position and x₀ is the initial position consistent with h and h₀ in mg, ALWAYS!
 
  • #19
vela
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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
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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
vela
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Go back and read post #9. I went over the different cases there. In particular, look at the last two examples.
 
  • #22
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What I did said:
½(12kg)(v² - 0) + 12(9.8)(0 - 6) + [tex]W = (5\hat{i}~\mathrm{N})\cdot(-11\hat{i}~\mathrm{m}) = (5~\mathrm{N})(-11~\mathrm{m}) = -55~\mathrm{J}[/tex] = 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
vela
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How did you derive this equation?

½(12kg)(v² - 0) + 12(9.8)(0 - 6) + W = 0
 
  • #24
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It's just this ΔKE + ΔPE + ΔTE = 0
 
  • #25
vela
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Reread the first paragraph of post 9.
 

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