Oh boy I think I have really messed up now :redface:
The way I think of it:
Energy Expanded by the weight - Friction opposing motion = Final Energy exerted on the Car
a.k.a
Energy Expanded by the weight = Final Energy exerted on the Car + Friction opposing motion
Thanks for your...
Work Done - Friction = Change In Energy
mgh - \muFn = mgh
x*9.81*.250 - .01?(.840+x)*9.81 = (.840+x)*9.81*D
2.4525x - .01?(.840+x)*9.81 / ( (.840+x)*9.81 ) = D
That is a recipricol, and theoretically there should be a point where there is a optimum weight, where the car will travel...
Ahh of course thanks :smile: .
Now, the only problem is, when I integrate friction into the formula
Work Done - Friction = Change in Energy
And re-arrange for distance with respect to x, I end up with a recipricol. This is no help, as there will be a point that the mass is optimum for...
Homework Statement
Hey guys, a gravity car is driven by a weight connected to a string which in turn is connected to the axle, which the weight drives when dropped. I believe the solution is very simple, however I just cannot grasp it. My task is to calculate the distance of a gravity car...