00:01
Okay, so in this problem we're going to use newton's role and kinetic of particle to solve this problem.
00:12
Okay, so first thing we need to know is that when the two newton of force applied to block a, okay? there will be an acceleration, we call it its acceleration south of a, equal.
00:35
4 we know it equal to f .a.
00:42
Divided by n .a.
00:45
By plugging the number, we have 2 newton.
00:52
The mass of block a is 0 .225, okay, kilogram.
01:11
Okay, so which gives a acceleration of 8 .89 meter per second.
01:31
So the same for b we have fb divided by n b equal this is 5 nton and this is 0 .6 kilogram okay so which get us a acceleration of 8 .3 meter per second square.
02:19
So that's the first step.
02:22
So the next step is to calculate the displacement during the time when the force is applied on these two blocks.
02:42
Okay, so by calculate the displacement, we use the equation of s .a.
02:52
Equal to 1 half, okay, delta t squared.
03:04
Because the initial velocity, in this case, for both blocks, is zero.
03:12
So we can use this equation like this.
03:14
Okay, by plugging the number, we have one half.
03:20
This is a point a nine.
03:25
The delta t equal to 0 .1 second.
03:29
So 0 .1 square.
03:35
Okay.
03:36
So in this case it gets get 0 .04 meter.
03:52
Same as block b, we need to calculate the displacement of block b during the force its prime.
04:07
So we use the same equation and the acceleration of block b is 8 .33 and times 0 .1 square and it will give us a value of 0 .42 meter.
04:53
Okay.
04:55
And then, okay, so then we know, in the beginning, these two blocks, it's 3 .4 meter apart...