00:01
Hello, here we have to solve the given problem and first we have to find acceleration of the upper block.
00:07
So before we proceed to solving this problem, let's analyze the system.
00:13
So let's presume that if the friction coefficient between m2 and m1 is very high, then block m2 won't be sliding with respect to the earth, with respect to the surface.
01:07
Meanwhile here obviously this coefficient is not high enough and that's why m2 is moving.
01:17
And with respect to this surface m2 is moving to the right.
01:28
That's why its acceleration a2 is directed to the right.
01:33
Now let's find the forces which are exerting on blocks m1 and m2.
01:40
Let's start with the block m1.
01:43
Block m1 is exerting force f, reaction n1, which is directed upwards, gravity mg1, and set two friction forces.
02:12
So the friction force between the surface and between the top block, so let's label them f friction 1 and f friction 2.
02:23
So let us presume that f friction 1 is a friction between the surface and block 1 and f friction 2 is a friction between block 1 and block 2.
02:58
And besides of and he also here there will be two gravity forces m1g and m2.
03:10
And overall that equals to m2a1 so now let's look at the forces which are exerting on block 1 sorry on block 2 which is on top this block is exerting gravity and reaction of this of the block 1 which is n2 additionally due to the fact that there is a surface which is pulling which resists the motion of the first block f2 according to the third newton's law the same force will be acting on the second block but directed to the opposite side and due to the same reason the force and two will be act the force of negative and 2 will be acting on the second on the first block by by the second block so therefore here we also have this component negative n1 and that is sorry negative and 2 and that is taken in the count as m2g so overall for the second block the second newton law equation can be written as following.
04:57
So f, christian 2 plus n2 plus n2g equals to m2a.
05:09
And that is the equation which we have to solve in question one to calculate acceleration of the second book.
05:20
Let's introduce x and y axis, x to the right, y to the top.
05:31
And let's project this.
05:34
Second newton's low equation for the second block on the x -axis.
05:39
So f -f2 equals to m -2a.
05:48
And projection onto y -axis gives us that n -2 equals to m -2g.
05:59
Additionally, f -f2 equals to mu, the coefficient of friction between the blocks, which is 0 .2, then this is coefficient is kinetic.
06:17
We have a set of three equations and we can combine them into one equation.
06:31
Let's not forget that this is acceleration of the second block and therefore a 2 equals to muqa 2 times g.
06:42
So that equals to 0 .2 multiplied by 9 .8 meters per second squared.
07:00
Let's calculate that is 1 .96 meters per second squared which is roughly 2 .0 meters per second squared.
07:15
So that's answer to the first problem.
07:19
Yeah.
07:20
So basically it's similar to yours except the sign because the first block, the second block will be moving to the right.
07:31
Now let's solve the second problem.
07:34
And in the second problem, we have to calculate the acceleration of the lower block...