- #1
Cosmophile
- 111
- 2
Homework Statement
This is not a homework or coursework question for myself. A friend sent it to me and I'm simply looking for assistance in mathematically proving my answer to myself. Here is the question:
"A cart rolls across a table two meters in length. Half of the length of the table is covered with felt, which slows the cart with a constant acceleration. Where should the felt be placed so that the cart crosses the table in the least amount of time?"
Homework Equations
Frankly, I'm having a difficult time finding the a footing here. I know that the frictional force ##\vec {F_f} = \mu \vec{F_n}##. It's obvious that we are wanting to maximize the average velocity (thus reducing the total time for the trip), but I'm not sure how to express this mathematically, or even if that is the best approach. I have also considered attempting to minimize the work done by friction.
The Attempt at a Solution
I know that ##W = \Delta K_E = \frac {1}{2} m{v_2}^2 - \frac {1}{2}m{v_1}^2 = \frac {1}{2}m(\Delta v)##
Because the cart eventually comes to a rest, the above equation can be simplifed to say
[tex] W = -\frac {1}{2}m{v_1}^2 [/tex]
I also know that, for an inclined plane, ##F_a = mg\sin\theta##, ##F_N = mg\cos\theta##, ##F_f = \mu F_n = \mu mg\cos\theta##
I also know that the work done by friction ##W_f = F_f \bullet d##, where ##d## is the displacement of the block. This gives [tex] W_f = \mu mg\cos\theta \bullet d = |\mu mg\cos\theta||d|\cos\phi [/tex], where ##\phi## is the angle between ##F_f## and ##d##. Because the two are in the same direction, ##\cos \phi = 1##, thus
[tex] W_f = \mu mgd\cos\theta =-\frac {1}{2}m{v_1}^2 [/tex]
[tex] W_f = \mu gd\cos\theta = -\frac{1}{2}{v_1}^2 [/tex]
And here is where I hit my wall.
Another friend has told me that approaching this via the Work-Energy principle is a dead-end and that I should instead approach the problem as a pure kinematics problem as opposed to dynamics. That being said, I'm exhausted and need to rest. Any help is greatly appreciated, as always.