00:01
For this problem, i think a good place to start is by redoing the graph, or the figure rather, but drawing it as a free body diagram to make it a little bit more clear what's happening.
00:15
So if this is our object that's being pushed, we know that it's feeling a friction force down the slope that is labeled the force of friction.
00:26
It has some weight directed downward, mg.
00:34
And we're actually given the x and y components of that force.
00:40
So we're told that the component that is along the slope is 0 .64 times mg, and that the component that is perpendicular to the slope is 0 .7.
00:58
7mg.
01:00
We are also given a normal force.
01:05
So i'm going to draw this direction because that's what makes more sense to mean.
01:10
I'm going to label that f sub n.
01:13
And lastly, we have the force that's being applied, which i'm going to draw in here.
01:19
It has a magnitude of 500 newtons.
01:22
But we are again given the components that are along and perpendicular to the slope, we're told the component that's along the slope is 383 newtons, and that the perpendicular component is 321 newtons.
01:42
And so because all of these are a little bit skewed, i'm going to go ahead and skew my coordinate axes as well.
01:53
So i'm going to say the x direction is up the slope, and the y direction is out of.
02:00
This slope for the positive y direction and so in order to figure this out we need to utilize newton's second law in both the x direction and the y direction so let's start with newton's second law in the x direction in that case we know the sum of the forces in the x direction must be equal to mass times acceleration because we're told that you accelerate in the x direction with an acceleration of 0 .75 meters per second squared up the slope and our mass is 25 kilograms.
02:42
But we can also add up the forces in the x direction.
02:46
So again, remember the x direction is along the slope with positive forces being up the slope and negative forces being down the slope...