00:02
So in this question we are given that there is a ramp shown with angle 30 degrees and there's a spring that is at the end of the ramp and so it's it lies in an unstretched length equilibrium length and there's a box that is four meters away sorry that is four meters away from the spring of mass m box of mass m such that we can write the values there is a spring constant k as well.
00:40
So the mass m of the box is 10 kgis and the stiffness of the spring is 200 newton per meter.
00:49
So we are told that there is no friction between the box and the inclined plane.
00:53
It's a frictionless inclined plane and the box is released from top.
00:57
So what would be the maximum compression of the spring when the box comes in, hits the spring.
01:04
So in this case what will happen is that the box will move and reach a position like this.
01:10
It will have some kinetic energy and then it doesn't stop here it this energy will let can get converted into so first potential energy gets converted to kinetic energy of the box and then the kinetic energy gets converted into further moves a little bit more from the equilibrium position let's let's call that extra distance as x that would be the compression of the spring and that would also mean that the spring is getting compressed by x and also that the box is also more moving x after it hits the spring as well.
01:46
So the total distance in the box moves along the inclined plane is 4 plus x.
01:57
So that would be the total distance it moves.
01:59
So we can directly say that that would be the potential net.
02:02
Let's say it's from center of the box to center of the box.
02:06
This height h is the height that it moves down in terms of potential energy vertically down because the gravity is acting vertically down.
02:17
So with respect to the work done against gravity.
02:19
Or by the gravity on the box you can say is related to mg times h that the potential energy loss in the energy of the box.
02:28
So we can say that the mg h of the box gets converted into half kx square of the spring which is the potential energy of the spring.
02:41
Simply we can bypass the step where we will first find the kinetic energy of the box and then let that further divide into loss of potential energy at distance x and then the spring constant based pernation energy spring we can directly say this so from here we can say that this is m g is 10 times 9 .81 times 4 plus x so h is d sine 30 sign 30 degrees because this is again 30 degrees because parallel to the angle that we have already measured so 4 plus x is the hypotenuse distance and h is sine 30 of that.
03:30
So we have to do 4 plus x sine 30 is equal to half k is 200 times x square.
03:39
So we can drop the half on both sides, sine 30 is half.
03:42
So this half goes.
03:43
And what we get is if we rearrange this a little bit, we get a quadratic equation, which is 200x square minus 98 .1x minus 38 .1x minus 3 .5.
03:56
392 .4 is equal to 0.
03:59
So you bring all the x -n constant terms on the right side.
04:03
So you have positive coefficient x -square term...