00:01
Hi there, so for this problem, we are told that starting from rest at 5 kilograms block.
00:07
So let's put in here the mass m is 5 kilograms.
00:12
And it's lies 2 .5 meters down a row 30 degrees incline.
00:21
So the distance that we're going to call d is equal to 2 .5 meters.
00:26
And the angle that we're going to call theta is equal to 30 degrees.
00:33
Now, the coefficient of kinetic friction between the block and the incline is also given that coefficient of kinetic friction is equal to 0 .436.
00:47
And so we need to determine for part a of this problem the work done by the force of gravity.
00:55
So to see how this work is, let's first draw this problem, the condition that we have for this problem.
01:05
So let me just start with the incline surface.
01:11
Something like this.
01:14
This is the incline.
01:18
Okay, something like this.
01:20
And we have on top of this a block.
01:24
So let's put the block in here.
01:29
Okay.
01:30
So, well, the draw is a little ugly, but it will give you the idea.
01:39
Okay.
01:40
So we are going to have the war.
01:48
Well, so the first question is the war done by the gravity, by the force of gravity.
01:54
So we know that gravity always points towards the center of the earth.
01:58
So this is the force of gravity.
02:01
Remember that the gravity is defined as the mass, the acceleration due to gravity.
02:06
Now, the displacement of this block is down to this slide.
02:11
So this is the displacement vector.
02:15
Now, the angle that the force, which in this case is the weight, meets with the vertical is this angle theta, which is the same angle in here.
02:29
So, as you can see, the angle that the force meets with the displacement, in this case, is going to be 90 degrees minus theta.
02:38
So that angle, 90 degrees, minus theta.
02:42
Minus theta, remember that theta is equal to 30 degrees.
02:46
So that will give us an angle of 60 degrees.
02:50
So this is the angle between the force vector and the displacement vector, because we know that the work done is equal to the force, the cross, the dot period between the force and the displacement.
03:03
Now, we can also write that as the magnitude of the force times the magnitude of the displacement times the cosine of the angle between these two.
03:11
So in this case, it's 60 degrees.
03:13
So the force is the mass times acceleration due to the gravity.
03:18
The displacement is the distance d that we are given, and this times the cosine of 60 degrees.
03:23
So we just need to simply substitute all of those numerical values.
03:28
So the mass that we are given is 5 kilograms.
03:32
The acceleration due to gravity is 9 .8 meters per second square.
03:36
This times the distance d that we are given, which is 2 .5 meters.
03:42
And this times the cosine of the angle which is the cosine of 60 degrees so using our calculator we obtain a value of 61 .25 yes 61 .25 jules so that's a solution for part a of this problem now for part we are asked about the word done by the friction force between the block and the incline.
04:32
Now, the frictional force always opposes the direction of motion of the block.
04:41
Now, as you can see, the angle between the frictional force and the displacement vector is equal to 180 degrees.
04:50
So, in the case f and b, we have that the work done by the frictional force, let's put an r in here to the free mc8.
04:59
So that is the frictional force that product with the displacement.
05:03
So in this case is the magnitude of the frictional force times the displacement times the cosine...