# Mousetrap Car Problem

## 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

Last edited:

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

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
Homework Helper
Gold Member
2020 Award
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.
You need some way to find the total energy available.
Where does the 0.94N figure come from?

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

hutchphd
Homework Helper
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.

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
Homework Helper
Gold Member
2020 Award
I know that the potential energy in the spring is 0.2044. Is that useful?
Yes indeed. Consider the work done against friction.

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

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
Homework Helper
Gold Member
2020 Award
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.

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

Yes indeed. Consider the work done against friction.
Thanks for the help.

Orodruin
Staff Emeritus