00:01
High so we are given that a mass fm is equal to 0 .8 kg is held against a spring whose spring constant is 56 n per meter compressing it by the distance of d meters as shown in the figure.
00:26
So when it is released from the rest it slides along the horizontal surface and this first encounter the graph of length l is equal to 0 .4 meter which coefficient of kinetic friction mu k is equal to 0 .25 and then at point p as shown so the mass encounters the smooth circular vertical path with the radius of r is equal to 0 .8 meter while value of g we have to let 10 meter per second square it is given in the question.
01:00
So in the e part we have to compute the value of theta so let get started with solution.
01:07
So what is theta here? so basically what about the circular path discovery from this point p to that of the point q this were indicated in the diagram as angle theta which basically we have to compute.
01:23
So let's say this is nothing but what the length r.
01:25
So at this point p basically what is happening the kinetic energy given to the block by spring.
01:34
So we can say that the kinetic energy given to block by spring this can be given by k is equal to 1 half k into d square and when block reach at the point p it would have lost 0 .25 times of its initial kinetic energy as per the question.
02:01
So remaining on 0 .75 percent of this kinetic energy 1 half k d square.
02:10
So this should be equals to 0 .75 into 1 half into k is given 56 times of d square this will comes to 21 d square.
02:22
This is what the remaining one remaining kinetic energy.
02:28
Now next point we need to note at point q the block has reached up to a vertical height of vertical height of r minus r cosine of theta.
02:45
So this should be equals to the r 1 minus of cosine theta but at the same point potential energy at point q that can be given as mg into height.
03:03
So height is what r 1 minus of cosine theta.
03:08
So let's plug in the values mass is 0 .8 g stain r is 0 .8 1 minus cosine of theta as we have to compute.
03:19
Now let the potential energy at q be u.
03:23
So thus at q at q clearly we can say the potential energy plus the kinetic energy this should be equals to the total energy according to the law of conservation of energy.
03:41
So we are having all the required values just equate them that is potential energy 6 .41 minus cosine of theta plus kinetic energy 1 half into 0 .8 v square...