# Slope With Varying Friction

1. Oct 29, 2014

### BOAS

1. The problem statement, all variables and given/known data
On a winter’s day in Maine, a warehouse worker is shoving boxes up a rough plank inclined
at an angle α above the horizontal. The plank is partially covered with ice, with more ice
near the bottom of the plank than near the top, so that the coefficient of friction increases
with the distance x along the plank: $µ = Ax$, where $A$ is a positive constant and the
bottom of the plank is at $x = 0 m$. (For this plank the coefficients of kinetic and static
friction are equal: $µ_{k} = µ_{s} = µ.$) The worker shoves a box up the plank so that it leaves
the bottom of the plank moving at speed $v_{0}$.
Show that when the box first comes to rest, it will remain at rest if $v_{0}^{2} \geq \frac{3g \sin^{2} (\alpha)}{A \cos (\alpha)}$

2. Relevant equations

3. The attempt at a solution

Essentially, unless the box has large enough initial velocity, it will not reach far enough up the plank so that the static friction prevents it from sliding back down. It is therefore necessary for $f_{s} \geq mg \sin (\alpha)$, the component of the box's weight down the slope.

$f_{s} \geq \mu_{s} N \geq xAN$

$x \geq \frac{mg \sin (\alpha)}{AN}$

$x \geq \frac{mg \sin (\alpha)}{Amg \cos (\alpha)}$

It feels like this is a relevant step, but i'm struggling to see where to go from here.

I don't know how to relate this to initial velocity...

2. Oct 29, 2014

### jbriggs444

Try conservation of energy. Equate the kinetic energy lost to gravity and kinetic friction to the starting kinetic energy.

3. Oct 29, 2014

### OldEngr63

Pretty hard to see how energy can be conserved in a system with friction.

jbriggs444, did you perhaps mean to suggest work and energy?

4. Oct 29, 2014

### BOAS

I'm confused by the distinction you are making...

Can I not solve this problem by saying KE = GPE + Wfriction?

I have managed to reach an equation that resembles what i'm trying to show, but I can't get rid of the Δh term from GPE, which is making me thing that this is the wrong approach.

I have reached $v^{2} \geq 2g \Delta h + \frac{2g \sin^2 (\alpha)}{A \cos (\alpha)}$

5. Oct 29, 2014

### OldEngr63

For energy to be conserved, there must be no energy lost (this is what conserved means). The equation written by the OP looks like a statement of work and energy, where the work against friction is included.

6. Oct 29, 2014

### haruspex

jbriggs444 is right. Energy is always conserved in a closed system, but some work can be lost as heat energy.

7. Oct 29, 2014

### haruspex

8. Oct 29, 2014

### Dr.D

@haruspex

What is there to indicate that this is a closed system? Friction work is energy lost to this mechanical system, so by definition, energy is not conserved.

9. Oct 29, 2014

### haruspex

You can always make a system closed by including enough. In this case, the box, the plank and the Earth as a mass should do it.
No, work done against friction is work lost to this mechanical system, so work is not conserved. Energy includes e.g. heat energy and is always conserved in a closed system. http://en.wikipedia.org/wiki/Conservation_of_energy

10. Oct 30, 2014

### OldEngr63

@ haruspex:

What an absurd, unuseful answer! No wonder you are a Science Advisor. You will never help the OP understand the difference between conservative and nonconservative systems.

11. Oct 30, 2014

### OldEngr63

@ haruspex:

From the viewpoint that you espouse, all systems become conservative and thus the term "conservative" becomes meaningless since nothing is excluded.

Actually, the "closed system" that you propose is not conservative. Some tiny amount of heat will be radiated away into interstellar space and is thus lost from your "closed system." There is no way to make such a closed system that will be conservative without including the entire cosmos. Otherwise, some amount of heat will always escape the system by radiation.

There was nothing in the original problem statement to suggest that the entire earth should be included in the system boundary. For a system comprised of the plank and the box, heat energy does escape the system and is thus lost. No larger system definition is meaningful in this problem. How can you not see that?

12. Oct 30, 2014

### jbriggs444

We should shelve the argument and work on the problem at hand. There is an energy budget and we can write an equation for it. "Conservation of Energy" may not be the best epithet to describe the source of that equation, but the equation is what matters.

I've worked through the problem based on that approach and it appears to lead directly to the intended solution.

The equation that I derive based on the energy budget

$\frac{mv^2}{2} = KE = mgsin(\alpha)x + \frac{mgAcos(\alpha)x^2}{2}$

Where x is "how far will the object get if launched at a particular velocity"

13. Oct 30, 2014

### OldEngr63

14. Oct 30, 2014

### haruspex

I'm sorry you find it unhelpful. I am at pains here to ensure the OP is not misled about the distinction between work and energy.
The standard distinction as I understand it (in physics) is that there are various forms of energy, but heat energy is a bit special (though arguably sound energy should be included there) in that there are constraints on how much can be converted to other forms. The more useful forms are collectively known as work energy, and may be further discriminated as mechanical work, etc.
Conservation of energy is a physical law (modified a bit by GR, I believe). The total energy in a closed system does not change. In the present case, we can reasonably ignore any heat lost from the plank and box system to the wider universe, and I only included the earth as a gravitational mass because it is required for the consideration of the potential energy (but we can finesse that by treating it as a field of irrelevant origin).
Non-conservation arises where we choose to neglect some forms of energy, or movement of energy outside the system. In the present context, work is not conserved because we do not count heat as work.
Can you agree with the bulk of the above? If not, let's take this offline, and perhaps involve some experts, and not risk confusing BOAS any further.