00:01
In this problem, we're given that there are two boxes and they are connected by a light inextensible string, which are as shown.
00:09
And the boxes a and b, these are released from rest.
00:14
So the mass of this box a, that's 12 kg, and the mass of this box b, well, that's 3 .8 kg.
00:23
And these boxes as they are released from rest, it is observed that box b, it falls through a distance of 0 .5 meters, and we will have to consider the coefficient of kinetic friction between the surface in contact for this block a and the horizontal surface.
00:44
That's 0 .2, and we are required to determine the speed of these two boxes when b has 4 .5 .5 .5.
00:53
Down by 0 .5 meters and we need to use here conservation of energy.
00:59
So we see that from this work energy theorem, the network that will be done on these boxes that has to be equal to the change in kinetic energy.
01:08
And we see that the network that's done here, it includes the work done by gravity on box b and the work done by friction on box a.
01:18
So this will be equal to the total final energy of the system minus the the total initial energy of the system.
01:26
And well, the system is released from rest.
01:28
So if we look at the equation for kinetic energy as its half mv square, the total initial kinetic energy that has to be zero...