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# Weight powered car

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we want the given distance each car travels is the same , equal

The work energy and distance traveled for each car being equal would mean the total friction is equal (regardless of how fast they go or how far they coast or anything else)

In your example friction is 0.05N so both cars would go d=W/f (=.98 joules / .05 N) 19.6 meters.

The purpose of this lab, however, is to get the cars to go as far as possible—not to get them all to go the same distance. As such, I'm curious what you're trying to calculate above.

~modest

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I gave the friction for Car K.

I am trying to figure out if two slightly different built cars can go the same distance Hopefully 10meters, at different speeds.

The purpose of this lab, however, is to get the cars to go as far as possible—not to get them all to go the same distance. As such, I'm curious what you're trying to calculate above.

The OP was just asking some simple questions to help them pass a simple task given to them at school. We turned it into entry level engineering.. :(

I'm looking for relationship and insight into the physics of this. Some of which has been overlooked imo, and got some facepalms in my efforts. I with drawl that statement as it is not accurate, and basically just stirs the pot for squabbling.

I aim for the path of understanding, and we are movin along.

A peice of my own advice to remind myself of:

"Save your worries and your whining for your dying breaths...nothing is over here... we've got work to do."

:shrug:

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I gave the friction for Car K.

I am trying to figure out if two slightly different built cars can go the same distance Hopefully 10meters, at different speeds.

I see. I think, yes, you can accomplish what you're looking for. The distance traveled is:

$d=\frac{w}{f}$

So if two different cars are going 10 meters and they both have 0.98 Joules energy then they both will need an average friction of 0.098 Newtons. There's no reason a fast and a slow car could not both have that average friction. The fast car would get it more from wind resistance and the slow car would have to make up for that with wheel friction or axle friction. But, yeah, I agree with you it certainly can happen.

~modest

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Okay.. Im quite certain I've got those source equations correct and ready to use.

We will say:

we know:

- we want the given distance each car travels is the same , equal

- the force for each car to accelerate (9.8N)

we need to find:

- the time it takes each car

- mass of each car

- the friction of each car

- acceleration of each car

- velocity of each car (at point where it enters coasting mode)

- the energy of each car

So in order to accomplish this, We need to give some characteristics to one car, and fill in the rest for the other car based on the first car data.

Car "K" will be known Car "U" will be unknown.

K is:

mass= 1kg + 1kg mass = 2kg

friction = 0.05N

Acceleration Force = 9.8N

There that should suffice, to begin solving.

Okay. I can do this. I am going to try and fill out the details of a hypothetical car with significantly higher mass (than the other car) and significantly lower friction (than the other car).

Begin.

Plan:

First Step:

-Set Mass -car.

-Set Friction (neglect air).

-Set work distance.

-Set coast d (distance).

Second Step:

-Solve velocity initial -coast d.

-Solve average deceleration -coast d.

-Solve Kinetic Energy initial -coast d.

-Solve work -coast d.

Third Step:

-Solve acceleration avg -work d.

-Solve velocity final -work d (given as velocity initial -coast d)

-Solve Kinetic Energy final -work d.

-Solve Work -work d.

-Solve Power -work d.

Work Path:

First Step:

-Set Mass -car. = 2kg

-Set Friction (neglect air). = 0.05N

-Set coast d (distance). = 2 meters

Second Step:

-Solve velocity initial -coast d.

(note: need deceleration first)

$a = \frac {F}{M} = \left(-0.025m/s^2\right)$

$a = \frac {0.05}{2} = \left(-0.025m/s^2\right)$

$\downarrow$

$v = a \, t = \left(-0.025m/s^2\right) \, \left(t)$ ? what is t

note: find t through velocity

$\downarrow$

Note: find hypothetical needed velocity to coast 2 meters @ M and friction. need to start finding energy, move into velocity needed to acquire such energy, solve t by given velocity:

$W = {f}{d} = \left(-0.05N \right) \times \left(2m \right) = 0.1J$

$\downarrow$

note: move onto energy

$Ke = \frac {1}{2} mv^2$

$v = \left ( \sqrt { \frac {2Ke}{M}} \right)$ (added after first pass)

solve V

$\downarrow$

$0.1 = \frac {1}{2} 2 v^2$

$\downarrow$

$0.1 = \frac {1}{2} 2 v^2$

$\downarrow$

$(0.1)(2) = 2 v^2$

$\downarrow$

$\frac{(0.1)(2)}{2} = v^2$

$\downarrow$

$\frac{(0.1)(2)}{2} = v^2$

$\downarrow$

$\sqrt{(0.1)} = v$

$\downarrow$

$v = 0.1m/s$

$\downarrow$

note: solve t

$v = \frac {d}{t}$

$\downarrow$

$t = \frac {d}{v}$

$\downarrow$

$t = \frac {2m}{0.1m/s}$

$\downarrow$

$t = 20$

note: 20 seconds unrealistic. Boosting fricton should resolve unrealistic time. (should of noticed that when acceleration output was 0.1m/s :lol:, awfully slow)

Trying: 2N friction (repeating same process above, solving work 1st)

$\downarrow$

W = 0.4J

V = 0.4m/s

t = 5seconds

note (5 seconds for 2meters sounds reasonable to apply to a realistic mouse trap car situation). Hmm could choose time first, and work it through, but I will stick with 5s.

(repasting 2nd step for easy reference)

Second Step:

-Solve velocity initial -coast d.

-Solve average deceleration -coast d.

-Solve Kinetic Energy initial -coast d.

-Solve work -coast d.

Things known:

W = 0.4J

V = 0.4m/s

t = 5seconds

Ke = 0.4J

d = 2m

$a = \frac {_dv}{ _dt} = 0.2m/s^2$

solving this strictly through latex as opposed to written hand on paper its utterly retarded... especially when you don't remember hardly any equations and have to manually swap stuff around and show work in latex.. lol.. FUN... :hyper: oh but wait.. I think I fixed the error.. EDIT... might be able to accomplish this.

(not quitting, continuing).

repeat,

Things known:

W = 0.4J

V = 0.4m/s

t = 5seconds

Ke = 0.4J

d = 2m

a = 0.2m/s^2

Second Step:

-Solve velocity initial -coast d. check

-Solve average deceleration -coast d. check

-Solve Kinetic Energy initial -coast d. check

-Solve work -coast d. check

Second Step Complete. :hyper:

(assuming I have at least 5 errors (typical of my physics work)

Third Step:

-Solve acceleration avg -work d.

-Solve velocity final -work d (given as velocity initial -coast d)

-Solve Kinetic Energy final -work d.

-Solve Work -work d.

-Solve Power -work d.

(entering 3rd step with some experience this time :phones: )

(moving work down in view)

W = 0.4J

V = 0.4m/s

t = 5seconds

Ke = 0.4J

d = 2m

a = 0.2m/s^2

-Solve acceleration avg -work d.

$a = \frac {_dv}{ _dt}$

hmm lacking needed data... might need to input a hypothetical.

okay, so max work output of trap car.

$W = Fd = (9.8N)(0.1m) =0.98J$

(checking to make sure things are good.)

$\downarrow$

$v = \left ( \sqrt { \frac {2Ke}{M}} \right)$

$\downarrow$

$v = \left ( \sqrt { \frac {(2)(0.98)}{2}} \right)$

$\downarrow$

$v = 0.9899m/s$

Oh... I've realized a mistake.

Ke is not 0.4j at .4m/s of a 2kg moving object.

I squared decimals.. and thats wrong isnt it?

If I use velocity as 40cm/s and then convert back to meters I get

Ke = 16j @ 0.4m/s

Okay.. i've been at this too long.. and I am getting in over my head.. I dont know if you can or can not square or square root decimal numbers in this situation.....

I submit to help... :confused:

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Morning Arkain. Bit early for math, it'n it? I submit to help... :confused:

I'll give it my level-best.

First Step:

-Set Mass -car.

-Set Friction (neglect air).

-Set work distance.

-Set coast d (distance).

Are you wanting to set the acceleration and deceleration distance and solve max velocity? That's possible. Let's see...

First Step:

-Set Mass -car. = 2kg

-Set Friction (neglect air). = 0.05N

-Set coast d (distance). = 2 meters

This will be a problem. If you want the total distance traveled to be 10 meters (which I think you said earlier) then the resistance on the car can't be 0.05 N. We can solve what it needs to be fairly easily by considering that the car starts its trek with zero kinetic energy and ends with zero kinetic energy. That will mean *all* the potential energy of the weight will go toward friction no matter what. That means total work = total distance times force of friction (w=d•f).. if friction is constant.

We know the potential energy of the weight is 0.98 Joules and we know total distance from start to stop is 10 meters, so the force of friction is w/d (0.98/10) 0.098 Newtons. So, that's the friction you should be using instead of 0.05 (if you want the total distance to be 10 m).

If I judge your intent correctly you're wanting to know the top velocity given some arbitrary deceleration distance (you say 2 meters). We know that the only force acting on the car for that 2 meters is friction which is known. Knowing force and distance we can solve for work (work = force x distance).

$w = f \times d$

$w = 0.098 \ N \times 2 \ m$

$w = 0.196 \ J$

If that's how much work it takes to stop the car (and it must be given the constraints we've introduced) then that's how much kinetic energy the car has when the weight has completely dropped and it's ready to start decelerating. We now have everything we need to solve for the max velocity:

$KE = \frac{1}{2}mv^2$

$v = \sqrt{2\frac{KE}{m}}$

$v = \sqrt{2\frac{0.196}{2}}$

$v = 0.4427 \ m/s$

You also might want to solve what the acceleration is on the first leg of the trip (the accelerating part). Knowing the velocity at the start (0) and the velocity when it's done accelerating (0.4427) and the distance over which it accelerates (8 m) we can use a simple kinematic relationship to solve for acceleration:

$v^2 = v_0^2+2ad$

omitting [imath]x_0[/imath] since it's zero and rearranging to solve for a:

$a = \frac{v^2}{2d}$

$a = \frac{0.4427^2}{2 \times 8}$

$a = \frac{0.4427^2}{2 \times 8}$

$a = 0.01225 \ m/s^2$

And, that's the acceleration on the first leg of the trip (the accelerating part). This provides us with an opportunity to use acceleration to solve for force (the force needed to accelerate the car in this manner) which will then allow us to check our answer. With acceleration and mass we solve for force:

$f=m \times a$

$f=2 \times 0.01225$

$f=0.0245 N$

The resistance of fiction is already subtracted from this figure. Now we know the force and the total work for the acceleration leg of the trip meaning we can solve for distance:

$d = \frac{w}{f}$

$d = \frac{0.196}{.0245}$

$d = 8 m$

And that's correct. It accelerates for 8 meters then decelerates for 2. What this boils down to is: if your car has friction of .098 N and you wanted it to accelerate for 8 meters then decelerate for 2 you would want to gear the car such that the force of propulsion is equal to 0.1225 Newtons (a rather slow-moving car). Subtract friction from that number and you have the force accelerating the car (0.0245 N)

Is this anywhere near what you're thinking Arkain? It is... you know, early and all. :eek_big:

~modest

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EDIT

Yeah, that is much easier to work out when you give the first leg a given distance. This was something I was trying to avoid, because the experiment I was doing forced it to be an unknown.

Oh wait, I think I misread... i will have to look over this agan.

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It occurs to me that, if you’re allowed to stretch the definition of “car” to “something with wheels”, you could take advantage of this one’s need to have no mechanical attachments other than gravity (unlike a mousetrap car), to minimize mass, and eliminate bearing, and pulley friction by eliminating the chassis altogether. Though it was too much bother to sketch it, the wheels could be lightened by removing most of their materiel, as in the picture in post #47.

The design sketched in the attached image is nothing but 2 wheels of about 30 cm diameter joined by an axle, a drive string connected to the weight and wrapped around the axle, and a support string looped loosely around the axle and attached to the weight to prevent it from descending more than the allowed 10 cm and touching the ground. The axle has a diameter less than 1/100th that of a wheel, so it should roll more than 10 m under power, coasting perhaps a bit further. ##### Share on other sites

I agree. I don't think it could get more efficient than that—no axle or pulley friction at all. No weight from a chassis. Quite ingenious. EDIT

Yeah, that is much easier to work out when you give the first leg a given distance. This was something I was trying to avoid, because the experiment I was doing forced it to be an unknown.

Oh wait, I think I misread... i will have to look over this agan.

You could also start with an arbitrary propulsion force (basically my post above in reverse). Here's a table of data in case it'll help. It solves different acceleration/deceleration (i.e. max velocity) profiles given the constraints that 1)the total distance is 10 meters and 2)friction is constant (all units are SI):

You can play with the figures and formulas if you have excel. The file is attached.

To forgo the assumption that friction is constant with velocity we'd need to use calculus as work is the area under a force curve—it's integral.

~modest

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It occurs to me that, if you’re allowed to stretch the definition of “car” to “something with wheels”, you could take advantage of this one’s need to have no mechanical attachments other than gravity (unlike a mousetrap car), to minimize mass, and eliminate bearing, and pulley friction by eliminating the chassis altogether.

Simply beautimous!!! ##### Share on other sites

• 3 weeks later...

The design sketched in the attached image is nothing but 2 wheels of about 30 cm diameter joined by an axle, a drive string connected to the weight and wrapped around the axle, and a support string looped loosely around the axle and attached to the weight to prevent it from descending more than the allowed 10 cm and touching the ground. The axle has a diameter less than 1/100th that of a wheel, so it should roll more than 10 m under power, coasting perhaps a bit further.

I can't even figure out how to get the attachment to display ( turtle down!), but nonetheless something about this particular elegant design keeps nagging me. With 30cm wheels, the radius is 15cm, and subtracting the 10cm drop we have but 5cm left. Allowing say a 1mm clearance between ground and fully fallen weight, we have only a 4.9cm clearance between the top of the weight and the axle when the weight is fully up. The upshot is that these exact design dimensions constrain the physical dimensions of the 1kg weight to a degrree that a "standard" lab 1kg weight may not fit. :doh:

The thinner the weight in its vertical dimension, the smaller diameter wheels one can use down to a limit of maybe 10.2cm (20.4cm diameter with a 1mm clearance & 1mm thick weight), but making the weight thin, one must either use a longer axle and separate the wheels wider, or have the weight stick out front & back, or both. Given these new constraints, what might make the ideal situation now for a 2-wheel 1kg weight-powered vehicle? (Shall we specify a lead weight for arguments sake? Gold? ):smart: :steering:

PS just realized no reason the weight could not have a slot for the axle & string to pass through & actually start & remain partially above the level of the axle at the end of the fall. :naughty:

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I can't even figure out how to get the attachment to display ( turtle down!),

In most browsers (IE, Firefox, etc) you should just be able to click on the thumbnail, and open a larger view of it. Rightclicking on that should allow you to save the full size image – and see just how sloppy a mouse sketch it is. ;)

With 30cm wheels, the radius is 15cm, and subtracting the 10cm drop we have but 5cm left. Allowing say a 1mm clearance between ground and fully fallen weight, we have only a 4.9cm clearance between the top of the weight and the axle when the weight is fully up. The upshot is that these exact design dimensions constrain the physical dimensions of the 1kg weight to a degrree that a "standard" lab 1kg weight may not fit. :doh:
You’re right – with my quick estimated dimensions, a typical weight likely won’t fit.

A 1 kg lead sphere has a diameter of about $2 \sqrt{\frac{1000 \cdot 3}{11.34 \cdot 4 \cdot \pi}} \dot= 5.52 \,\mbox{cm}$, so even ignoring whatever’s used to attach it to the strings, it’ll hit the ground before descending 10 cm. :(

Rather than messing with odd-shaped weights, however, I think you could just use larger wheels – 40 cm diameter, for example. The axle, then, should be about 4 mm (or less, if possible) in diameter, and separate the wheels by 8 cm or so – enough to assure the mass doesn’t scrape either wheel if the vehicle rocks slightly, and as the string unwinds.

Reducing mass by cutting out as much wheel material as possible without compromising its strength is even more important with these bigger wheels.

The assignment/contest rules, I think, require the vehicle to have 4 wheels. I think you could satisfy this with a pair of small, light wheels attached to the axle with piano wire or similar, so that the vehicle remains effectively two-wheeled.

These school (10th grade, in my local public schools) tech assignment seem like great fun. There was not such class when I was in those grades. You could do stuff like this independently, but getting school credit for it would have been extra nifty. ;)

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• 7 months later...

Hello forum. First post so be gentle! With CraigD's idea what calculations will need to be produced to show the concept works?

Also am I correct in thinking the wheel width will have to be as minimal as possible to reduce friction?

(Linked image as JPG is quite large)

Just to add I have been assigned a similar project to the OP - instead of starting a new thread I thought I'd give this one a bump as my project is very similar.

If required I can post a copy of the project requirements on here. ## Join the conversation

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