00:01
Hi, here in this given problem there is a horizontal table top, a frictionless table top.
00:09
A block of mass m1 kept over 8, an ideal pulley attached at the edge of the table, an ideal string, holding the another block of mass m2 at its other end.
00:26
Weight of this block, m2g, acting vertically downward, tension t in the string, up.
00:32
Upward and the same tension t will be here also.
00:37
The system will be moving down the block of mass m2 will be moving down.
00:41
Suppose with the acceleration a, so m1 will be moving towards right with the same acceleration a.
00:48
Masses of the blocks m1 is equal to 3 .0 kilogram and m2 is equal to 2 .0 kilogram.
00:59
In the first part of the problem we have to find acceleration in the system.
01:05
So first of all we use free body diagram of the block of mass m2 as it is moving down.
01:19
So m2g will be more than tangent t and net force will be equal to m2 into a using newtons a second of motion.
01:29
And using free body diagram of the block of mass m1, there is only a single force acting on it t.
01:41
So that should be equal to m1 into a.
01:43
So putting this value of tension here in the first equation we get m2g minus m1a is equal to m2a or we get an expression for acceleration m2g divided by m1 plus m2 means that is 2 times of 9 .8 8 divided by 2 plus 3...