- #1
JackV
- 8
- 0
- 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
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
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