00:01
In this problem, we're driving in our car and we are locking our brakes so that the only force acting on the car in the x direction is the force of friction, which is slowing us down.
00:11
So the complete force diagram looks something like this.
00:15
If we apply newton's second law in the y direction, we know it's equal to zero, since the acceleration is equal to zero.
00:21
It's not going to jump up in the air.
00:23
This yields immediately that the normal force is equal to m times g.
00:27
Now, i know the friction force is equal to the coefficient of kinetic friction times normal force.
00:32
And we just found that the normal force was mg.
00:34
So we have that the friction force looks like this.
00:38
Now we can apply newton second law in the x direction.
00:41
In this case, we can't set equal to zero since we're accelerating in the x direction.
00:46
In this case, we get negative mute k, mg, or friction force here.
00:51
It's negative because i define my coordinate system like this, and it's moving in the, or the forces in the negative extraction.
00:59
It's equal to mass times the acceleration in the extraction.
01:02
The mass cancels out.
01:03
And we, after plugging in, get that the acceleration in the extraction is negative 7 .84.
01:09
And it should be negative because since the force of friction is in the negative, we're going to be accelerating in the negative.
01:16
Now that we have the acceleration, we can use a kinematic equation in order to solve for the distance travel.
01:22
We know that the initial velocity is 28 .7...