00:01
All right, so we have two blocks on a table.
00:04
The bottom block is we'll call block b, and then on top of that is block a, and they're connected via a string that runs over a pulley, and block b is pulled with a force of f, and we're told the acceleration of each block is 1 .95 meters per second squared.
00:23
The coefficient of friction between block a and block b is 0 .132 .3.
00:30
We don't need the static coefficient or friction because on this problem, they're already moving.
00:34
So we'll just use the kinetic coefficient.
00:37
And let's see, we're told that the mass of block a is 1 .88 kilograms.
00:44
The mass of block b is 2 .1 kilograms.
00:50
And then lastly, the coefficient between b and the table is basically one -fourth.
00:56
So let's draw a free body diagram of block a.
00:59
Because what we want to find is the force, f, and then the tension in the string connecting the blocks.
01:03
But we'll actually do this in the reverse order.
01:05
So for block a, we're going to have the tension going to the left and the frictional force on block a going to the left.
01:12
Tension going to the right, excuse me.
01:15
So newton's second law for this one says like m .a.
01:18
Times a equals t minus f .a.
01:23
And f, the frictional force here, is just going to be the coefficient of friction between the two blocks times the massive block a times its weight.
01:31
So we can see the tension then is going to be like m .a.
01:37
Times a plus mu k times g...