00:01
Hi, here in this given problem first of all this is the horizontal table top the rough horizontal table top a block kept over it then this is a pulley mass less pulley mass less frictionless pulley mass less thread passing over it joining the two blocks.
00:28
Suppose this is block a this is block b mass of block b that is 6 .00 kilogram.
00:35
So, its weight 6 .00 g will be acting vertically down tension t in the string in both of the segments of the string about this pulley the same tension.
00:48
Here this block a is having a mass 8 kg.
00:53
So, its weight 8 .00 g acting vertically down.
00:57
So, normal reaction n acting up which will be equal to 8 .00 g using newton's third of motion as the block a will be moving towards right.
01:09
So, force of friction kinetic friction will be acting towards left.
01:15
Initial speed of both of the blocks that is 0 .900 meter per second and finally, the system comes to rest.
01:23
So, v f will become 0 after moving through after covering a distance of 2 .00 meter.
01:32
So, starting from starting with block b for block b using work energy theorem work done on this block b that will be equal to the force acting on it which is 6 g downward minus tension t upward...