00:01
So we have this block with a weight, which i will call fg because we're going to be using work and let's not get confused with that stuff.
00:08
A weight of 40 newtons and it sits on a frictionless incline plane.
00:15
So like this.
00:21
And we know that this incline plane makes an angle of 42 degrees with the ground.
00:31
And we have some applied force acting parallel to the incline.
00:37
And up the incline.
00:38
So we have some applied force f, and f is equal to 26 .8 newtons.
00:51
And so it is sufficient to pull the block up the plane at a constant speed.
00:56
This is super important.
00:59
We're going to be moving up, so we're going to moving up the incline at some constant speed.
01:05
And we also have these other forces.
01:07
We have normal force acting up.
01:09
We have the y component.
01:13
Of weight acting down, and we have the x component of weight acting down the incline.
01:23
And so we want to know for part a, what is network on this object? well, we know that network is equal to the change in kinetic energy, but kinetic energy changes based off of the speed changing, and is the speed change? changing? no.
01:49
So we know that network for this block moving up by a distance of l equals 6 meters.
01:59
It will go from here to here, let's say.
02:02
We know that network must equal 0 joules because the speed of the block is not changing, because it's moving at a constant speed.
02:11
And we'll see that this holds true.
02:14
So in part b, we want to know the work due.
02:19
To gravity.
02:22
So the work due to gravity will be equal to the work due to the x component of gravity plus the work due to the y component of gravity.
02:34
And so we first need to realize what is the formula for work? the formula for work is equal to force times the distance traveled or the displacement traveled times cosine of theta where theta is the angle between your displacement vector right here x and whatever force vector we are using so for gravity we have the x component of weight force times that distance l times cosine of what so our displacement vector is parallel to to the surface of the incline and is going up.
03:24
Our force vector for the x component of weight is parallel to the incline, but it's pointing down.
03:30
So the angle between the two is 180 degrees, plus force of gravity in the y direction times l times cosine of what angle.
03:44
Well, our displacement is pointing up the ramp and our y component of weight force is pointing perpendicular to the ramp...