answer a financial math question

[giblets]

Hell, why not
i) M(S^2)/2
ii) M(S^2)/18
iii) M(S^2)/9
iv) 4M(S^2)/9
v) M(S^2)/9
vi) M(S^2)/3
M is teh mass of the slab, S is its initial speed
Did I screw up?

[Solomwi]

I'd be absolutely shocked if I'm right and the two of you are wrong, but here's my work:

Fn = 2M*g, the normal force at the frictional surface only countering the weight of the small block
Fk = Mu*Fn = Mu*2M*g = 2M*(Mu*g)
a = Mu*g
vf = vi(small) + a*t = vi(large) - a*t --> 0 + Mu*g*t = S - Mu*g*t
--> S = 2*Mu*g*t --> t = S/(2*Mu*g)

Then, substituting for t in the original vf equation for either block yields S/2:

Large:
vf = S - Mu*g*t = S - Mu*g*S/(2*Mu*g) = S - S/2 = S/2

Small:
vf = 0 + Mu*g*t = Mu*g*S/(2*Mu*g) = S/2

Edit: now that I look at it again, my solution doesn't appear to depend on the relative masses at all, probably a pretty good sign that it's wrong.

[giblets]

Solomwi, one object is twice the mass of the other, and they are experiencing the same force from friction, so the lighter object will have twice as much acceleration as the heavier one. The slab is losing speed at 2µg, while the block is gaining at µg, until they reach equal speed, I think.

[KrazyHorse]

Implying that I've gotten a basic physics problem wrong is generally not a good idea, but I'll let it slide.

[Hauldren Collider]

But the bottom line here is basically that Ben is a ****ing idiot. Although we knew that anyway.

[Solomwi]

Solomwi, one object is twice the mass of the other, and they are experiencing the same force from friction, so the lighter object will have twice as much acceleration as the heavier one. The slab is losing speed at 2µg, while the block is gaining at µg, until they reach equal speed, I think.

Yep, that's what I get for making a slapdash run at it. Stupid mistake, but at least I know I had the concepts right.

KH:

If that's the inference you took, you may well be cut out for law . The intended implication was that I had probably overlooked something, and I showed my work to get help finding it.

[KrazyHorse]

I wasn't taking offence. You're generally reasonable in your intellectual honesty. Just playing around a bit.

[Solomwi]

I'm impeccable in my intellectual honesty, dammit, not merely generally reasonable.

[Solomwi]

But the bottom line here is basically that Ben is a ****ing idiot. Although we knew that anyway.

I have to wonder if, had you originally given the energy questions too (particularly the one about waste heat), Ben would have used the same conservation of energy approach to the ones you did give. I also have to guess yes.

[Ecofarm]

What if uh C-A-T really spelled dog?

[Hauldren Collider]

I have to wonder if, had you originally given the energy questions too (particularly the one about waste heat), Ben would have used the same conservation of energy approach to the ones you did give. I also have to guess yes.

I think it's plausible you could try to hack this apart using energy equations and stuff but I don't feel like thinking about that. The only real "trick" here (which is only a trick if you're a beginner) is observing Newton's Third Law--the kinetic friction applies an equal and opposite force to the slab.

[Solomwi]

I think it's plausible you could try to hack this apart using energy equations and stuff but I don't feel like thinking about that. The only real "trick" here (which is only a trick if you're a beginner) is observing Newton's Third Law--the kinetic friction applies an equal and opposite force to the slab.

With the information given, I don't know that you could quantify (or in this case, express in some combination of S, M, Mu, and g) the waste heat other than from the first law of thermo as the difference between initial and final kinetic energy, which in turn requires knowing vf.

Eki(slab) = Ekf(slab) + Ekf(block) + Q
M*S^2/2 = 3M*vf^2/2 + Q

Anyway, the point is I doubt even waste heat being explicitly mentioned would have deterred Ben from jumping right into the same fundamental mistake he wound up making.

[Hauldren Collider]

Hours later I'm still getting a huge kick out of the fact that he can't solve a problem that I did during the first week of this class. Even with his amazing 2 years of physics education.

[Riesstiu IV]

I just find it hard to believe at this point that Ben is a real person. It has to be some elaborate troll.

[giblets]

HC, maybe you can answer a question that Kuci got wrong.

Betelgeuse is a star located 640 ly away from the earth.

RA is 5h 55m

Dec is +7 deg 24'

What are the coordinates of betelgeuse in cartesian coordinates, assuming the Sun is 0/0/0

Since I'm bored I decided to look up "RA" (which turns out to be "right ascension") and "declination" on wikipedia and attempt this thing.

5h 55m = 5(15) +55(.25) = 75 + 13.75 = 88.75º angle
7º 24' = 7 + 24/60 = 7.4º angle

declination seems related to teh z-axis in Cartesian coordinates so I'll start with that. I'll assume earth and the sun are in the same place because eight light minutes isn't going to make a difference. so.... if I imagine a right triangle, with the hypotenuse being the 640 ly from earth to Betelgeuse, the side opposite of our 7.4º angle is going to give us the location on teh z-axis. So..... sin(7.4º) * 640 ly = ~82.4 ly

Now, right ascensions seems related to teh x and y axes, and the other leg of my right triangle, the side adjacent to my 7.4º angle. Which is around 634.7 ly. I need to break that into x and y. one of these will be around 634.5 ly and the other will be around 13.8 ly. Since I used sine to get the big one I'll say that one's the distance on the y axis, and the one I used cosine to get can be the x axis. It could go either way I guess, and I don't know anything about astronomy so I don't know what people normally do.

So my coordinates... (x,y,z) : (13.8,634.5,82.4)

Is that about right?