# Gravity Car Distance

1. Aug 23, 2011

### PhyzWizKid

1. The problem statement, all variables and given/known data
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 given the weight of 840g (plus the falling weight, x) and the distance it falls. Initial velocity is 0m/s.The distance will be with respect to x, being the falling weight powering the car.
Mass of car = 840g
Mass of weight = x
Distance Weight Falls = 250mm
Distance = ???
2. The attempt at a solution
Well, the work that the falling weight will do is =mgh =.250*9.81*x = 2.4525x
The mass of the car =.840+x
Friction =u*Fn =.01*840+x*9.81

Distance =???

Thanks guys.

2. Aug 23, 2011

### tiny-tim

Welcome to PF!

Hi PhyzWizKid! Welcome to PF!

(have a mu: µ )
(you mean .01*(.840+x)*9.81 )

ok so far

now use the work energy theorem: https://www.physicsforums.com/library.php?do=view_item&itemid=75" = change in energy …

what is the relation between https://www.physicsforums.com/library.php?do=view_item&itemid=39" and work done?

Last edited by a moderator: Apr 26, 2017
3. Aug 23, 2011

### PhyzWizKid

But with the work energy theorem, change in energy requires a velocity doesnt it? (EK=1/2mV^2)
Thanks

4. Aug 24, 2011

### tiny-tim

Last edited by a moderator: Apr 26, 2017
5. Aug 24, 2011

### PhyzWizKid

Ahh of course thanks .
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 greatest distance (a peak in the data, but a recipricol does not have this)?

Thanks.

6. Aug 24, 2011

### tiny-tim

I don't understand.

What is your equation?

7. Aug 24, 2011

### PhyzWizKid

Work Done - Friction = Change In Energy
mgh - $\mu$Fn = 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 farthest (quadratic), beacuse as the pushing force increases, so does its resistance (total system weight)?

Cheers

8. Aug 24, 2011

### tiny-tim

eugh! :yuck:

you have no distance in your https://www.physicsforums.com/library.php?do=view_item&itemid=75"

and why do you have mgh on both sides?

Last edited by a moderator: Apr 26, 2017
9. Aug 24, 2011

### PhyzWizKid

Oh boy I think I have really messed up now
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 time

10. Aug 24, 2011

### tiny-tim

times distance
This I don't understand.

Work done = ∆KE + ∆PE.

∆KE = 0, so there are only two non-zero terms in the equation …

what is your "Final Energy exerted on the Car" ?​

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