Register to reply 
Friction stopping a car 
Share this thread: 
#1
Mar208, 04:38 PM

P: 2

1. The problem statement, all variables and given/known data
Given two cars, one heavier than the other but identical in all other respects, traveling at the same speed on a snowy or icy road which one will stop sooner? Momentum would say that the lighter car should stop sooner but how does the heavier car's increased weight in combination with friction figure into the stopping distance? Shouldn't the heavier car be able to dig into snow more increasing the opposing frictional force? On ice even though neither car can "dig in", wouldn't the heavier car experience a greater frictional force causing it to stop sooner or is the increased momentum greater than this increase friction? Explanations are fine as are equations. Thanks for the help. 


#2
Mar208, 04:53 PM

P: 86

One dependency is how much the angular momentum at the wheels is shifted into linear velocity. . .
p.s. this is a true driver dependency. 


#3
Mar208, 04:59 PM

P: 2

How about I simplify the problem.
Let's just say neither car has ABS and to stop both drivers choose to mash on the brakes and lock up the wheels. So essentially both cars are just sliding blocks. 


#4
Mar208, 05:22 PM

P: 86

Friction stopping a car
Frictional force is independent of weight, but dependent on material composition.
Momemtum is mass * velocity. p = mv. Friction will retard speed. Friction uses energy directly. Kinetic energy(KE) is (m * v * v)/2. So let us say KE = mK for this example. Friction will leave r*m*k units of energy to continue the work. With one mass m1 and the other m2, the KE ratio, yields m1/m2 as the deciding factor of who travels the furthest. Example, if both cars are identical, then m1 = m2, and they both stop at the same time. This makes sense because you know heavier cars use more gas to attain the same speeds as lighter cars and since friction is independent of weight, the solution is complete. The important point is at the beginning of this post. 


#5
Mar208, 09:35 PM

P: 456

where k is the coefficient of friction. If you combine that with F=m.a to compute the deceleration of the car, you'll see that THAT doesn't depend on mass. 


Register to reply 
Related Discussions  
Stopping distance considering coefficient of friction  Introductory Physics Homework  6  
Stopping a car.  Advanced Physics Homework  7  
Friction and stopping  General Physics  5  
Friction and stopping distance  Introductory Physics Homework  5  
Stopping the car  Introductory Physics Homework  2 