00:01
Hello, here a mass m is sent down an incline with an initial speed of v0.
00:05
So let's first show this diagram because in the first question we have to draw the free body diagram showing all the forces.
00:14
So let's presume that this is an incline with an angle theta.
00:21
And this is the box.
00:25
The box is exerting gravity, which is mg reaction, normal reaction of the incline.
00:34
And also the kinetic friction.
00:43
And the friction is parallel to this surface.
00:47
So that is a free body diagram.
00:51
And there, these three forces results in acceleration.
00:59
Now, so that's answered question a.
01:05
Now we have to answer question b.
01:07
So here we first have to find the component of box weight acting down the slope.
01:13
Let's introduce y and x -axis and let's find the first component.
01:27
So here we have to find mgx and this is this component.
01:36
Angle teta is this guy, oh sorry, angle teta is this angle.
01:50
Then mgx equals to mg times.
01:58
Times sine teta.
02:01
And now let's find the force of friction.
02:05
The force of friction equals to mu times n and n equals to m g cosine alpha.
02:18
Then this force of friction equals to mu k times m g cosine alpha.
02:34
Now let's answer question c.
02:37
Here we have to find work.
02:39
And in the first sub -question, we have to find the work done by the friction as the box slides down all the way the incline.
02:51
So the length of the incline is d.
02:54
Therefore, work done by the friction force equals to negative mulek m g.
03:01
Cosine alpha times d.
03:04
And this is negative because the direction of f is...