00:01
For this problem on the topic of force and motion, we are given two blocks in the figure, 1 and 2, with masses 2 kg and 1 kg respectively, and they are connected by a string of negligible mass.
00:12
Block 2 is pushed by a force f, which has a magnitude of 20 newton's in a direction of 35 degrees.
00:18
We want to find the coefficient of kinetic friction between each block, or we are given the coefficient of kinetic friction between each block to be 0 .2, and we want to find the tension in the string.
00:30
So if m2 is the 1kg block, we can apply newton second law, and we get in the x direction f cosine theta minus the tension t minus the kinetic frictional force fk is equal to m2 times a.
00:52
In the y direction, for block two we have the normal force fn minus the vertical component of f, which is f sine theta, minus the weight of block 2, which is m2 times g, must equal to 0, since there's no motion, no net motion in the vertical direction.
01:14
Now we know fk is equal to mu k times fn, and so these two equations can be combined into a single equation...