What is the problem with the coefficient of friction in my mousetrap car design?

[JackV]

Homework Statement

I have a mousetrap car that goes for 6.366 meters on a frictionless track and weighs 2.35 Newtons. If there is 0.94 Newtons worth of friction force opposing it, how far would the car go now?

I was just wondering if anyone knows where to start on this.

Thanks.

Relevant Equations

Ffriction=u(coefficient of friction)*Fnormal

Fnet=m*a

Ffriction=0.94N
u (coefficient of friction)=0.4
Fnormal=2.35N
Mass of Cart= 0.24 kg
Spring Potential Energy of the Mousetrap=0.2044 J
Circumference of Wheels= 0.373 m
Circumference of Axles= 0.0047625 m
Drive Train Length= 0.254m
Theoretical Distance based on Circumference of Wheels, Circumference of Axles, and Drive Train Length Without Friction Taken Into Account: 6.366 meters

 

[hutchphd]

If it is on a frictionless track why does it stop at all?

 

[JackV]

If it is on a frictionless track why does it stop at all?

6.336 meters is the theoretical value that I calculated based on the length of the drivetrain and the circumference of the axle and wheels. So it is not really frictionless as much as theoretical stopping distance without friction taken into account.

 

[haruspex]

6.336 meters is the theoretical value that I calculated based on the length of the drivetrain and the circumference of the axle and wheels. So it is not really frictionless as much as theoretical stopping distance without friction taken into account.

Then it is still not useful in answering your question.
You need some way to find the total energy available.
Where does the 0.94N figure come from?

 

[JackV]

I know that the potential energy in the spring is 0.2044. Is that useful?

 

[hutchphd]

6.336 meters is the theoretical value that I calculated based on the length of the drivetrain and the circumference of the axle and wheels. So it is not really frictionless as much as theoretical stopping distance without friction taken into account.

Perhaps you should start again and give a more comprehensive description of what you have done and what you want to know.

 

[JackV]

Then it is still not useful in answering your question.
You need some way to find the total energy available.
Where does the 0.94N figure come from?

The 0.94N is from multiplying the normal force of the cart (2.35 N) and the coefficient of friction (0.4).

I know that the potential energy in the spring is 0.2044. Is that useful?

 

[haruspex]

I know that the potential energy in the spring is 0.2044. Is that useful?

Yes indeed. Consider the work done against friction.

 

[JackV]

Perhaps you should start again and give a more comprehensive description of what you have done and what you want to know.

Good idea.

 

[JackV]

Perhaps you should start again and give a more comprehensive description of what you have done and what you want to know.

I have now added all of the information that I have to the description.

 

[haruspex]

I have now added all of the information that I have to the description.

Ok, but the spring PE is still the place to start. See post #8.

 

[JackV]

Ok, but the spring PE is still the place to start. See post #8.

Thanks.

 

[JackV]

Yes indeed. Consider the work done against friction.

Thanks for the help.

 

[Orodruin]

I know that the potential energy in the spring is 0.2044. Is that useful?

No. Without units it is utterly useless.

 

[willem2]

The 0.94N is from multiplying the normal force of the cart (2.35 N) and the coefficient of friction (0.4).

a coefficient of friction 0.4 is an extremely high number. This is the highest value I can find on the web for "Car tires on loose sand". With so much friction the cart likely won't move at all.
You need rolling friction, not sliding friction, and you should be able to do more than 100 times better with the friction coefficient.