Static Friction and time and distance problems

[lucybug]

Hi I was having trouble with a few static questions from the Exam Krackers 1001 Physics book. Question number 226 and 227 talk about a car moving at a certain velocity given (Vi and Vf) and then brakes and slides to a stop. We take into account the coefficient of Kinetic Friction (given) and it ask for the distance and time.

I am not sure how they are coming up with the formulas to answer the question.

I know that if we have v(i) and v(f) as well as the acceleration we can use
vf^2=vi^2 + 2ax
But how do we factor in the coefficient of kinetic friction?

Thanks for the help.

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[Phlame217]

give me the specific problem and i can help.

 

[EkramVahsedi86]

A car moving at 20 m/s and slides to a stop. If

is 0.1, then how far does the car slide before it stops.

Well to solve this, you have to realize that "gravity" is what slows the car down. Gravity is 9.8 meters per second per second, and

is 0.1.

To understand why, realize that if gravity was 0 meters per second, then the car would never slow down even if the tires lock and the brakes lock.

20 meters per second is initial velocity
acceleration is the product of g and

Plugging in 20 div by .98 = 20.4 seconds

The average velocity during those 20.4 seconds is

1/2 of V_initial plus V_final

1/2 of (Vf + Vi) = 10 meters per second

During those 20.4 seconds of sliding, the car's average velocity is 10 meters per second.

20.4 times 10 = 204, or basically 200 meters.

This was one of the hardest problems on friction i have seen, so don't be at all discouraged. Just remember that

is the ratio of the frictional force to the normal force.

The normal force acting on the car is ma

F = ma

where the mass of the car is m
where the gravity is a

The normal force is therefore ma which is now m times 9.8

That was as far as I got without looking into my physics notes.

You now simply multiply 9.8m times 0.1 to get the frictional force

Friction = 0.98 * mass of car = 0.98 m
The momentum of the car is 20m
(We know the velocity of the car is 20 meters per second but we need to either drop the m term on Friction or add an m term to our car's velocity so that the two will cancel out).

Our car's initial momentum is 20m and it experiences a frictional force of 0.98m and when you divide 20m by 0.98m we get 20.40816327

The formula I used was momentum divided by force = time

Now we have time = 20.4 seconds (which you can round to just 20 seconds)

The car slides to a stop from 20 meters per second, and the whole slide takes 20 seconds. The easy formula now to use next is

D = V_avg * t
V_avg = average velocity = 1/2 (V_initial + V_final) = 10 m/s
Distance = V_avg * t = 10 m/s * 20 s = 200 m

 

[EkramVahsedi86]

A car moving at 20 m/s brakes and slides to a stop. If

mewk is 0.1, then how long does it take the car to stop

Oh tricky, they ask for the time when you need that for the first problem! How tricky and clever! Its very easy to find the time.

Initial velocity = 20 m/s
final velocity = zero
acceleration = -mewk*N
where N is the normal force acting on the car, 9.8m

When you have a MCAT problem that has a term like (m) instead of a number, you have to change a variable so that the (m)s will drop off and cancel out since my calculator doesn't do algebra, just adding, subtracting and mulitplying.

The velocity of the car is 20, and its mass is m.

Momentum = mv
the car's momentum is 20m

Momentum div by Force = time
because Force * time = Momentum

The tricky part is to see why or how to simplify the problem from a hard one into a problem that has "just the numbers left". For this one, the (m) term messed up everything and needed to cancel from both sides somehow.

The alternative answer is to memorize a huge formula:
distance = V_f - squared minus V_i squared all divided by 2a
where a is Vf - Vi div by time

Vi = 20
Vf = 00
g = 9.8
mewk = 0.1
a = -(mewk)(g)
d = ( Vi^2 - Vf^2 ) / 2(mewk)(g) = 400 / 2(0.1)(9.8) = 204.08 meters

As you can see, its not intuitive to plug terms into a huge formula. If you change is to a force-momentum problem, then the physics is easier than a friction-motion problem.

 

[EkramVahsedi86]

Hi I was having trouble with a few static questions from the Exam Krackers 1001 Physics book. Question number 226 and 227 talk about a car moving at a certain velocity given (Vi and Vf) and then brakes and slides to a stop. We take into account the coefficient of Kinetic Friction (given) and it ask for the distance and time.

I am not sure how they are coming up with the formulas to answer the question.

I know that if we have v(i) and v(f) as well as the acceleration we can use
vf^2=vi^2 + 2ax
But how do we factor in the coefficient of kinetic friction?

Thanks for the help.

The coeffecient of kinetic friction affects acceleration.

Acceleration = g*mewk

This is the short answer but I recommend to try and follow the long answers or pm me since the MCAT is about thinking on the fly and seeing proportionalities just from visualization. If you can't "see" that gravity affects the normal force, then you won't understand friction questions. Don't get discouraged, my weak area on physics is optics. Maybe you can help me on optics

 

[amikhchi]

I find it a bit easier to do it this way (might be the same, just simpler):

in order for the car to stop, the frictional force is acting in the opposite direction of motion

Fk = mg * mewk, g ~ 10, therfore Fk = m
so your car of mass m has a force acting on it, what's its accel? using Fk = m:
F = m*a -----> m = m * a, therefore a = 1
now use Vf^2 = Vi^2 + 2ad using Vf =0, Vi=20, a = 1
you end up with 400 = 2d, so 200m (keep in mind that since accel is opposite of direction of velocity, one fo them will initiall be negative so the signs end up cancelling)

your initial question: the coefficient of friction comes in because that's directly involved with the force that is slowing the car down so using the coefficient of friction, you get the force acting on the car, once you get that, you find the accel of the car, and once you know that the rest is plug/chug

 

[Fort]

These are all great solutions. Another way to solve this problem is using the "Work-Energy Theorem".

Start with W(net) = (delta)KE

The equation simplifies to

W = -(1/2)mv^2

We also know that the net work done on object is equal to the product of the applied force and the distance that it traveled.

We can now write:

W = -(1/2)mv^2 = Fd

Solving for d we have:

d = (1/2)mv^2/F *Negative sign vanishes because our force is negative.

Now, the net force done on this car is frictional force, which we know to be u(k)F(normal)

Now

d = (1/2)mv^2/[u(k)F(normal)] = (1/2)mv^2/[u(k)mg]

Aha! The m's finally cancel and it's now just a matter of plug and chug.

d = (1/2)v^2/[u(k)g]

d = (1/2)(20)^2/[(0.1)(9.8)]

d = 204 m

Lots of one dimensional motion problems can be solved using the Work Energy Theorem and Conservation of Mechanical Energy. You'll find they work out nicely!