00:01
For this problem on the topic of force and motion, we have two cars a and b, b is stopped at a red light along a road that is handed down a hill, and the slope of the hill is 12 degrees.
00:12
The cars were separated by a distance of 24 meters when the driver of a pressed the brakes of car a and put it into a slide.
00:23
The speed of car a at the onset of braking was 18 meters per second, and we want to know the speed at which car a, car a will hit car b if the coefficient of kinetic friction is 0 .6 for a dry day and 0 .1 for a wet road surface.
00:41
Now we'll apply newton's second law to the downhill direction.
00:45
And so the weight mg sine theta of car a minus the frictional force is equal to m times a.
00:57
And we know that the frictional force f is that due to kinetic friction, since the car is moving, and this is equal to the coefficient of kinetic correction mu k times the normal force fn, which we can write as mu k m g cosine cosine theta.
01:23
So for part a, we have mu k equal to 0 .6.
01:32
And from the newton second law equation, we have a to be g sine theta minus mu k cosine theta...