00:01
All right, so in part a, we need to calculate the acceleration of this box in two different experiments.
00:11
It has a mass of 8 kilograms.
00:13
An experiment number one, it goes 8 .22 meters over a time interval of 2 .8 seconds.
00:23
I'm going to start from rest, so it has an initial velocity of zero.
00:26
So now we do kinematics to it in order to find the acceleration.
00:33
You know, assuming constant acceleration.
00:43
So we know the initial velocity is zero, so that term disappears.
00:48
2 .x squared.
00:54
Plug in numbers.
00:57
That's 8 .22.
00:59
Down here is 2 .8 seconds.
01:02
Sorry, 2 .8 seconds.
01:05
Plug in the numbers, and then we get 2 .10 meters per square second.
01:11
And this will be supporting evidence for our answers.
01:14
So let's put a box around it.
01:16
Okay.
01:16
And then in experiments number two, i guess we're giving the books, sorry, the box is a slightly smaller push because it only goes 4 .20 meters over a time interval of two seconds.
01:32
And again, it began at rest.
01:34
So v .0 is going to be 0.
01:35
We just use the same equation.
01:39
So let's just skip ahead to that step.
01:42
So 2, delta x over delta t squared.
01:46
Plug in numbers.
01:51
And see that we suspiciously get an identical value of acceleration.
01:58
That's 2 .10 meters per square second.
02:02
So yes, the boxes decelerate at the same rate.
02:08
Because kinetic friction only depends on the normal force and a constant.
02:15
Cognetic, kinetic.
02:17
Cricion.
02:44
Excellent.
02:45
Okay, so we are going to need a new slide because this is a data problem.
02:50
We need a table.
02:51
So let's go ahead and start setting that up.
02:58
Great.
02:59
And then that's terrible.
03:05
Oh my gosh.
03:05
Okay.
03:07
There we go.
03:08
All right...