00:01
The diagram of the problem is shown.
00:02
So for the first case, actually the change in potential energy, because there is no friction, will be equal to the change in the elastic potential energy of the block, or changing potential energy of the block, must be equal to potential energy stored in the spring.
01:06
Actually, it is better to write only energy because in case of mass -reservation, spring we have only this is the stored energy and this is in the nature of potential energy.
01:16
So from here we get, so you can see from this diagram we can write m g l1 sine theta, sorry, small not small, capital m, g l1 sine theta that is the changing potential energy must be equal to half k x square where x is the compression of the spring.
01:52
So from here we are getting x equal to root over of twice capital n g l1 sine theta divided by k so this will be the compression for the first case now for the second case we are actually having a coefficient of friction so in that case let's say v is the velocity at the bottom of the inclined plane so from here we can say v is equal to 0 plus 2.
02:45
Now what is the acceleration in this case? it is 2.
02:48
G multiplied by sine theta minus mu k cost theta multiplied by l1 because l1 is the distance which is traveled by the block in the inclined plane...