Calculate Gravity Car Distance with 840g + x Weight

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The discussion revolves around calculating the distance a gravity car travels when powered by a falling weight. The car's mass is 840g, and the weight's mass is variable (x), with the weight falling a distance of 250mm. Participants discuss using the work-energy theorem to relate the work done by the weight to the energy lost due to friction. There is confusion regarding the integration of friction into the calculations, particularly about achieving an optimal weight for maximum distance. The conversation highlights the need for clarity in applying energy equations and understanding the relationship between work done and friction.
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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 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.
 
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Welcome to PF!

Hi PhyzWizKid! Welcome to PF! :smile:

(have a mu: µ :wink:)
PhyzWizKid said:
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.

(you mean .01*(.840+x)*9.81 :wink:)

ok so far :smile:

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? :wink:
 
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But with the work energy theorem, change in energy requires a velocity doesn't it? (EK=1/2mV^2)
Thanks
 
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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 greatest distance (a peak in the data, but a recipricol does not have this)?:confused::confused:

Thanks.
 
PhyzWizKid said:
Work Done - Friction = Change in Energy

I don't understand. :confused:

What is your equation?
 
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 farthest (quadratic), beacuse as the pushing force increases, so does its resistance (total system weight)?

Cheers
 
PhyzWizKid said:
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

eugh!

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? :confused:
 
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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 time :smile:
 
  • #10
PhyzWizKid said:
Friction opposing motion

times distance :wink:
Final Energy exerted on the Car

This I don't understand. :confused:

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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