shortest amount it takes to rest & max friction

[SaintJude]

The passage gives you a value of car's mass, 1000 kg, and coefficient of static friction, 0.6 & also that of kinetic friction, 0.2.

Q: The car is traveling at a velocity of 24 m/s along a level road. What is the shortest time it will take for the car to come to rest? (Assume that the car does not skid, and neglect air resistance.)



Explanation:

The question asks for the least time, which means that the static force of friction is at its maximum value

What's the reasoning behind this?

Thus the acceleration is found from F= ma= ustatic * N & then one uses v = vo + at

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[pm1]

The passage gives you a value of car's mass, 1000 kg, and coefficient of static friction, 0.6 & also that of kinetic friction, 0.2.

Q: The car is traveling at a velocity of 24 m/s along a level road. What is the shortest time it will take for the car to come to rest? (Assume that the car does not skid, and neglect air resistance.)



Explanation:



What's the reasoning behind this?

Thus the acceleration is found from F= ma= ustatic * N

The only force applied to the car in order to stop it will be friction. Thus, Fnet = Ffriction, hence ma=Ffriction. Note that this acceleration is actually deceleration.
Since they ask for the shortest time for the car to come to a stop we want a max a, in order to get a max a we use max friction.

If you want to stop an object from moving as quickly as possible, it makes sense to use the max force against its current movement, right?

 

[SaintJude]

The only force applied to the car in order to stop it will be friction. Thus, Fnet = Ffriction, hence ma=Ffriction. Note that this acceleration is actually deceleration.
Since they ask for the shortest time for the car to come to a stop we want a max a, in order to get a max a we use max friction.

If you want to stop an object from moving as quickly as possible, it makes sense to use the max force against its current movement, right?

But why not use kinetic friction? In this quantitative case it wouldn't make a difference as it will be the same numbers. But if there was an answer choice that asked you which value you needed from the choices below, why would you choose a? I mean the car is moving...

a. determine maximum static friction
b. determine kinetic friction.

 

[MrNeuro]

i use impulse....i think you get the same answer

Favg t = delta mv

delta mv = 1000(24) = 24000

Favg = total net constant force applied Ffriction = us(N) = 6000

t= 4 seconds

the whole F=ma route w/ kinematics seemed a little odd to me but i think you should still get the same answer

 

[MrNeuro]

But why not use kinetic friction? In this quantitative case it wouldn't make a difference as it will be the same numbers. But if there was an answer choice that asked you which value you needed from the choices below, why would you choose a? I mean the car is moving...

a. determine maximum static friction
b. determine kinetic friction.

question says the cars not sliding...meaning its us also remember that when you slam on the breaks on your car your tires are actually still rolling thats why you have ABS (anti lock braking system) because as your wheels turn your tires are exhibiting static friction which exerts a larger deceleration than would a kinetic friction...

 

[SaintJude]

Ah, yes impulse is a great approach too.

I never understood why "Assume that the car does not skid" = indicates static friction. Why?

P.s. the passage was about "anti lock brake system..." but I just skimmed it..

 

[MrNeuro]

Ah, yes impulse is a great approach too.

I never understood why "Assume that the car does not skid" = indicates static friction. Why?

uk is the sliding coefficient

 

[ymartino]

You don't need to use impulse. You can use the equation F=us(N)=m(v/t). I'm replacing acceleration with (v/t) since we're looking for time. Using the coefficient of friction of 0.6*1000kg*9.81m/s^2, you get 5886=m(v/t). Multiplying mass by velocity gives 24000kgm/s. Then simply divide 24000kgm/s by 5886kgm/s^2 to get 4.1 seconds. Hope that helps!