Recent content by Bill M

I'm envisioning a more varying slope, with perhaps several "humps".

You're right that I originally asked only about the work done by friction. I originally was only asking about this component because I already knew how to find the work done by gravity (quite easy, as you point out). My end goal (though unstated) was to find the total work done and that's why...

Okay, see if this looks right: The sum of forces acting along the ground on the object would be: \sum F = w(\mu cos\theta + sin\theta ) where mu is the coefficient of kinetic friction and w is weight. This is equivalent to: \sum F = w(\frac{\mu +m}{\sqrt{m^{2}+1}}) where m is the slope...

If I were doing it for a straight surface, I'd use: W = mgfd, where m is mass of the object, d is the distance traveled, and f is an adjusted coefficient of friction based on slope (or angle theta). My problem now is that, first, distance is not so easily calculated (though I could use an...

Hello, I'm trying to figure out a method of calculating the work done by friction on an object sliding down a surface with a variable slope, assuming an equation can be determined to fit the line along which the object travels and we have a known coefficient of friction for the surface...

Thanks, I'll try simplifying a bit. 10 years out of practice, I guess I'm glad that I got an answer at all! Thanks for your input.

Yeah, if you wouldn't mind looking at it, I'd appreciate it. Like I said, it's been over 10 years since I've done this stuff! The square root in my solution came from the fact that r v2 was in the equation and that after solving for that, I just wanted v. You've made it simpler by...

Flip, V(x) is on the right side because the equation is in the form of forces = mass times acceleration. Since acceleration is V'(t) but I don't want it as a function of t, I want it as a function of position (x), I used these steps... a = dV/dt dV/dt = (dV/dx)(dx/dt) - chain rule...

Hey guys, I'm looking to calculate the initial velocity of a large object sliding to a stop over a surface. This is easy without the drag force of the air, but I'd like to include that to prove a point in my research. As a quick background, I haven't done a differential equation in over a...

Andrew, that's a good point. I suppose I'd like to ignore that for now and first figure out if I came up with the right solution: V(x) = \sqrt{-\frac{Mfg}{k}(1-e^{\frac{-2k(x-D)}{M}})} My calculations are showing that the drag produces very little difference in calculated speeds for a...

Okay, first a brief intro. I studied math/physics extensively as an undergrad...but that was over 10 years ago now. My day job keeps me doing more basic physics and math regularly, but I haven't, for example, solved a differential equation in over a decade! Anyway... I'm trying to...