00:02
So in this problem we are given that there is a 20 degree ramp on which there is a book, a physics book for that matter, which weighs 1 kgs and it's attached to a string that goes over a pulley and on the other side hangs a coffee mug which i'll just draw as a box.
00:29
And this coffee mug has a weight of 0 .5 kg or 500 grams.
00:34
And we are also told that the book and the inclined plane have coefficient of static friction as 0 .5 and the coefficient of kinetic friction as 0 .2.
00:51
And that the book is actually pushed in the direction up the inclined plane with the initial velocity of 3 meters per second.
01:02
So the first thing we have to find is how far does the book travel up the inclined plane before it comes to a stop.
01:11
So for that let's draw a free body diagram of the book.
01:14
So let's say this is the book and it has a weight.
01:19
Let's call this weight as m capital m and this weight as small m.
01:24
Just for the diagram sake.
01:26
So mg is the weight of the book acting downwards.
01:30
There is mg cost theta component where theta is 20.
01:36
This is theta and this would be the sine theta component.
01:47
Then there is a normal force.
01:48
From the inclined plane.
01:51
And since the body is moving up with some velocity we at any given point, let's say, before it comes to stop, but it's still moving up.
01:58
So friction always acts against the direction of motion.
02:02
So there will be a friction force acting downwards.
02:06
And this value would be mu k times fn because this would be kinetic friction because there is some relative motion between the book and the inclined plane.
02:16
And let's also assume that there is a, the book is decelerating.
02:20
So, so there is some relative motion between the book is decelerating.
02:20
So there is an acceleration in this direction because there is no force which is going to accelerate the body, the book towards the upside.
02:31
There are forces only reacting towards the bottom side of the inclined plane.
02:37
So there's most likely a desolation going to happen and that would create.
02:42
Since the desolation is happening, acceleration direction for desolation is towards the downward direction.
02:47
There will be a pseudo force which will be capital m times a, acting towards the top side.
02:54
So this is a little bit counterintuitive.
02:56
Although the body is moving upwards, it is decelerating.
03:01
So the pseudo -force acts upwards.
03:04
So these are the forces that act on the book.
03:07
And if we draw, and there's one more force.
03:09
I'm sorry, there's another tension force from this.
03:11
Let's say this string has tension t.
03:14
We will assume that this pulley has no mass and no moment of inertia so that we can say that the strings on both side have the same tension.
03:20
So another tension t would also be acting.
03:27
So for the coffee cup, let's say this is the tension t pulling up.
03:35
There is a small mg acting downwards.
03:39
And since in this case as well the acceleration is downwards, although the motion is upwards, but the acceleration is downwards because the book and the coffee mug have to accelerate in the same direction because the string is not an extendable string it's not a flexible string so in that case and it's a massless string as well so it cannot hold any pseudo force as well and create some acceleration of its own so the acceleration of the cup must match the acceleration of the book and so it will have a pseudo force upwards again which will be small m times a so with this we can write some equations balancing the forces in the directions given to us so for the book we can say in the direction that is perpendicular to the inclined plane, the fn force is balanced by mg cost theta.
04:34
So these are the only two forces in the direction.
04:37
So they balance each other out.
04:39
And in the direction that is along the inclined plane, we see that the forces acting down the inclined plane are t plus mg sine theta plus a mu k times fn, which we can put from here as mg cost theta.
04:54
And this is equated by the m times a.
04:59
So this is equation one.
05:01
And from the coffee cup diagram, we can say that t plus m times a is equal to m times g.
05:11
So we can substitute t from equation 2 into equation 1.
05:24
And when we do that, we get that t is equal to mg minus ma.
05:30
And we put that into equation 1.
05:31
So we get mg minus m a plus what we can do is we can start writing the values themselves.
05:42
So mg minus ma would mean 0 .5 times g minus 0 .5 times a plus so we have put t here.
05:53
We need to put these values now and that would give us 1 times g times sign of 20 plus us mu k which is 0 .2 times 1 times g cost of 20 is equal to 1 times a so we can take this 0 .5 a to the right side and simplify the because we if you put g equal to 9 .81 we know everything else on the left hand side so this gives us that 1 .5 times a is equal to or a is equal to 10 .08 divided by 1 .5 .5.
06:40
And this gives us the a acceleration is 7 .6 meters per second square.
06:45
So this is the acceleration due to or deceleration, let's say, in the downward direction.
06:51
So using this and the initial velocity which is which we are given, we can find out the distance covered by the book.
07:01
So let's do that...