00:01
Okay, so here you have a firefighter that's sliding down this pore here.
00:09
You're told that the poor in total is 4 meters long with this firefighter with about 75 kilograms.
00:22
And you're told that it's a 300 newton frictional force acting on this system.
00:31
And you're told that he lands on a platform that was that is 20 kilograms and that platform is supported by a spring that it has a spring constant k of 4 ,000 moutons per meter here so essentially you want first you want to calculate the velocity that it hits a platform with and then after that you want to see what kind of compression induces with that velocity so the first thing is first.
01:10
We want to find out how much speed this guy has when we get to the bottom.
01:17
So essentially this is a second low problem.
01:19
So you have the force of gravity minus this friction of force here.
01:32
This is so this is summing up all the forces that should give you the resultant force, which is mass -time acceleration.
01:39
So these are the two forces acting on this person as they slide down.
01:44
So you've noticed that this we can calculate this is just 75 times 9 .8 minus 300 and then you divide that by the mass of person which is just 75.
02:04
So once you do this you actually find the acceleration and the acceleration in this case is about 5 .8.
02:14
Meters for a square second and then what you want to calculate now is so once you have that now we want to find the velocity once they hit the bottom so we know that we don't we don't have anything in terms of time here so we want to use kinemite equations that don't involve time so the best one to use is that one that says vbano squared is equal to v initial squared plus plus 2a times of change in distance.
02:52
And in which case the change in distance here is just y.
02:55
So we know that the v initial here is zero because we're told that it starts from rest.
03:02
We assume it starts from rest if we are not told that.
03:05
So this a is what we just calculated here.
03:08
So that goes here and then this delta y is the four meters.
03:12
So we know that the fine velocity is going to be the square root of 2.
03:21
Two times the acceleration 5 .8 times the distance which is four and that should give us something like 6 .81 meters per second.
03:42
Then the more complicated about this problem is actually figuring out how much this thing drops back.
03:49
So we can't use kinematics in this in this scenario first, we have several different things going on...